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5.2: Solving Systems of Linear Equations Algebraically

Created by: CK-12

Introduction

In this lesson you will learn to solve systems of linear equations algebraically. When technology is not available, the solution to a consistent and independent system can be determined by using algebra. This is necessary if the solution is not integers.

The algebraic method that you will learn in this lesson is called substitution. A second method that you will learn is called the comparison method. This is actually a form of substitution.

Solving Systems of Linear Equations by Substitution

Objectives

The lesson objectives for Solving Systems of Linear Equations Algebraically are:

  • Determining the solution by substitution
  • Determining the solution by comparison

Introduction

In this lesson you will learn to solve a linear equation that has two variables by substitution. A 2 \times 2 system of linear equations consists of two equations with two variables. Substitution involves solving one of the equations in terms of one of the variables. This value for the solved variable will then be substituted into the other equation for the variable. The result will be an equation with just one variable. This equation will then be solved and you will have one part of the ordered pair. The value will then be substituted into the equation that was written in terms of one of the variables. This will result in the other part of the ordered pair. The solution has been determined by using the substitution method.

Another form of substitution is known as the comparison method. If two equations are written in terms of the same variable, they can be set equal to each other. The resulting equation can then be solved. The answer for this equation will then be substituted into one of the original equations to yield the value for that variable. The two answers are the ordered pair that solve the system of linear equations.

Watch This

Khan Academy Slightly More Complicated Equations

Guidance

When you solve a system of consistent and independent equations by graphing, a single ordered pair is the solution. The ordered pair satisfies both equations and the point is the intersection of the graphs of the linear equations. The coordinates of this point of intersection is not always integers. Therefore, some method has to be used to determine the values of the coordinates. The method that will be presented here is called the substitution method.

Solve by substitution:

\begin{Bmatrix}2x+3y=13 \\y=4x-5\end{Bmatrix}

The second equation is solved in terms of the variable ‘y’. The expression (4x+5) can be used to replace ‘y’ in the first equation.

2x+3y &= 13\\2x+3({\color{red}4x-5}) &= 13

The equation now has one variable. Apply the distributive property.

2x+{\color{red}12x-15}=13

Combine like terms to simplify the equation.

{\color{red}14x}-15=13

Solve the equation.

& 14x-15 {\color{red}+15} = 13 {\color{red}+15}\\& 14x = {\color{red}28}\\& \frac{14x}{{\color{red}14}} =\frac{28}{{\color{red}14}}\\& \frac{\cancel{14}x}{\cancel{14}} = \frac{\overset{{\color{red}2}}{\cancel{28}}}{\cancel{14}}\\& \boxed{x = 2}

To determine the value of ‘y’, substitute this value into the equation y=4x-5.

y &= 4x-5\\y &= 4({\color{red}2})-5

Evaluate the equation.

& \ y = {\color{red}8}-5\\& \boxed{y = 3}

The solution is (2, 3). This represents the intersection point of the lines if the equations were graphed on a Cartesian grid.

Example A

Solve the following system of linear equations by substitution:

\begin{Bmatrix}3x+y=1 \\2x+5y=18\end{Bmatrix}

To begin, solve one of the equations in terms of one of the variables. This step is simplified if one of the equations has one variable whose coefficient is either +1 or -1. In the above system the first equation has ‘y’ that has a coefficient of 1.

Solve for ‘y’.

& 3x+y = 1\\& 3x {\color{red}-3x}+y = 1 {\color{red}-3x}\\& \boxed{y = 1-3x}

Substitute (1-3x) into the second equation for ‘y’.

2x+5y &= 18\\2x+5({\color{red}1-3x}) &= 18

Apply the distributive property and solve the equation.

& 2x+{\color{red}5-15x} = 18\\& {\color{red}-13x}+5 = 18\\& -13x+5 {\color{red}-5} = 18 {\color{red}-5}\\& -13x = {\color{red}13}\\& \frac{-13x}{{\color{red}-13}} = \frac{13}{{\color{red}-13}}\\& \frac{\cancel{-13}x}{\cancel{-13}} = \frac{\overset{{\color{red}-1}}{\cancel{-13}}}{\cancel{-13}}\\& \boxed{x = -1}

Substitute -1 for x into the equation \boxed{y=1-3x}.

y &= 1-3x\\y &= 1-3({\color{red}-1})

Evaluate the equation.

& \ y = 1 {\color{red}+3}\\& \boxed{y = 4}

The solution is (-1, 4). This represents the intersection point of the lines if the equations were graphed on a Cartesian grid.

Another way to write ‘the lines intersect at (-1, 4)’ is:

\boxed{\text{Line1}} : 3x+y=1

\boxed{\text{Line2}} : 2x+5y=18

Line 1 intersects Line 2 at (-1, 4)

& \qquad \quad \text{at}\\& \qquad \quad {\color{red}\uparrow}\\& \boxed{l_1 \cap l_2 @ (-1,4)}\\& \quad \ {\color{red}\downarrow}\\& \text{intersects}

Example B

Solve the following system of linear equations by substitution:

\begin{Bmatrix}8x-3y=6 \\6x+12y=-24\end{Bmatrix}

There is no variable that has a coefficient of +1 or of -1. However, the second equation has coefficients and a constant that are multiples of 6. The second equation will be solved in terms of the variable ‘x’.

& 6x+12y =-24\\& 6x+12y {\color{red}-12y} = -24 {\color{red}-12y}\\& 6x = -24-12y\\& \frac{6x}{{\color{red}6}} = \frac{-24}{{\color{red}6}}-\frac{12y}{{\color{red}6}}\\& \frac{\cancel{6}x}{\cancel{6}} = \frac{\overset{{\color{red}-4}}{\cancel{-24}}}{\cancel{6}}-\frac{\overset{{\color{red}2}}{\cancel{12}y}}{\cancel{6}}\\& \boxed{x = -4-2y}

Substitute (-4-2y) into the first equation for ‘x’.

8x-3y &= 6\\8 ({\color{red}-4-2y})-3y &= 6

Apply the distributive property and solve the equation.

& {\color{red}-32-16y}-3y = 6\\& -32 {\color{red}-19y} = 6\\& -32 {\color{red}+32}-19y = 6 {\color{red}+32}\\& -19y = {\color{red}38}\\& \frac{-19y}{{\color{red}-19}} = \frac{38}{{\color{red}-19}}\\& \frac{\cancel{-19}y}{\cancel{-19}} = \frac{\overset{{\color{red}-2}}{\cancel{38}}}{\cancel{-19}}\\& \boxed{y =- 2}

Substitute -2 for y into the equation \boxed{x=-4-2y}

x &= -4-2y\\x &= -4-2 ({\color{red}-2})

Evaluate the equation.

& x = -4 {\color{red}+4}\\& \boxed{x = 0}\\& \boxed{l_1 \cap l_2 @ (0,-2)}

Example C

In this example, another form of substitution will be presented. This form is called the comparison method.

Jason, who is a real computer whiz, decided to set up his own server and to sell space on his computer so students could have their own web pages on the Internet. He devised two plans. One plan charges a base fee of $25.00 plus $.50 every month. His other plan has a base fee of $5.00 plus $1 per month.

i) Write an equation to represent each plan.

ii) Solve the system of equations.

Both plans deal with the cost of buying space from Jason’s server. The cost involves a base fee and a monthly rate. The equations for the plans are:

y=.50x+25 and y=1x+5 where ‘y’ represents the cost and ‘x’ represents the number of months.

Both equations are equal to ‘y’. Therefore, the equations are equal to each other.

\begin{Bmatrix}y=.50x+25 \\y=1x+5\end{Bmatrix}

& .50x+25 = 1x+5 && \text{Set the equations equal to each other and solve the equation.}\\& .50x+25 {\color{red}-25} = 1x+5 {\color{red}-25}\\& .50x = 1x {\color{red}-20}\\& .50x {\color{red}-1x} = 1x {\color{red}-1x}-20\\& {\color{red}-.50x} = -20\\& \frac{-.50x}{{\color{red}-.50}} = \frac{-20}{{\color{red}-.50}}\\& \frac{\cancel{-.50}x}{\cancel{-.50}} = \frac{\overset{{\color{red}40}}{\cancel{-20}}}{\cancel{-.50}}\\& \boxed{x = 40 \ months}

Since the equations were equal, the value for ‘x’ can be substituted into either of the original equations. The result will be the same.

y &= 1x+5\\y &= 1({\color{red}40})+5

Evaluate the equation.

& y = {\color{red}40}+5\\& \boxed{y = 45 \ dollars}\\& \boxed{l_1 \cap l_2 @ (40,45)}.

Vocabulary

Comparison Method
The comparison method is a form of the substitution method used for solving a system of linear equations algebraically. This method is used when both equations are equal to the same variable.
Substitution Method
The substitution method is a way of solving a system of linear equations algebraically. The substitution method involves solving an equation for a variable and substituting that expression into the other equation.

Guided Practice

1. Solve the following system of linear equations by comparison: \begin{Bmatrix}x=2y+1 \\x=4y-3\end{Bmatrix}

2. Solve the following system of linear equations by substitution: \begin{Bmatrix}2x+y=3 \\3x+2y=12\end{Bmatrix}.

3. Solve the following system of linear equations by substitution: \begin{Bmatrix}\frac{2}{5}m+\frac{3}{4}n=\frac{5}{2} \\-\frac{2}{3}m+\frac{1}{2}n=\frac{3}{4}\end{Bmatrix}

Answers

1. \begin{Bmatrix}x=2y+1 \\x=4y-3\end{Bmatrix}

Both equations are equal to the variable ‘x’. Set (2y+1)=(4y-3)

2y+1=4y-3

Solve the equation.

& 2y+1 {\color{red}-1} = 4y-3 {\color{red}-1}\\& 2y = 4y {\color{red}-4}\\& 2y {\color{red}-4y} = 4y {\color{red}-4y}-4\\& {\color{red}-2y} = -4\\& \frac{-2y}{{\color{red}-2}} = \frac{-4}{{\color{red}-2}}\\& \frac{\cancel{-2}y}{\cancel{-2}} = \frac{\overset{{\color{red}2}}{\cancel{-4}}}{\cancel{-2}}\\& \boxed{y = 2}

Substitute this value for ‘y’ into one of the original equations.

x &= 2y+1\\x &= 2({\color{red}2})+2

Evaluate the equation.

& \ x = {\color{red}4}+1\\& \boxed{x = 5}\\& \boxed{l_1 \cap l_2 @ (5,2)}

2. \begin{Bmatrix}2x+y=3 \\3x+2y=12\end{Bmatrix}

The first equation has the variable ‘y’ with a coefficient of 1. Solve the equation in terms of ‘y’.

& 2x+y = 3\\& 2x {\color{red}-2x}+y = 3 {\color{red}-2x}\\& \boxed{y = 3-2x}

Substitute (3-2x) into the second equation for ‘y’.

3x+2y &= 12\\3x+2 ({\color{red}3-2x}) &= 12

Apply the distributive property and solve the equation.

& 3x+ {\color{red}6-4x} = 12\\& 6 {\color{red}-x} = 12\\& 6 {\color{red}-6}-x = 12 {\color{red}-6}\\& -x = {\color{red}6}\\& \frac{-x}{{\color{red}-1}} = \frac{6}{{\color{red}-1}}\\& \frac{{\cancel{-1}}x}{{\cancel{-1}}} = \frac{\overset{{\color{red}-6}}{\cancel{6}}}{\cancel{-1}}\\& \boxed{x = -6}

Substitute this value for ‘x’ into the equation y=3-2x.

y &= 3-2x\\y &= 3-2({\color{red}-6})

Evaluate the equation.

& \ y = 3 {\color{red}+12}\\& \boxed{y = 15}\\& \boxed{l_1 \cap l_2 @ (-6,15)}

3. \begin{Bmatrix}\frac{2}{5}m+\frac{3}{4}n=\frac{5}{2} \\-\frac{2}{3}m+\frac{1}{2}n=\frac{3}{4}\end{Bmatrix}

Begin by multiplying each equation by the LCM of the denominators to simplify the system.

\frac{2}{5}m+\frac{3}{4}n=\frac{5}{2} The LCM for the denominators is 20.

{\color{red}20} \left(\frac{2}{5}\right)m+ {\color{red}20} \left(\frac{3}{4}\right)n &= {\color{red}20} \left(\frac{5}{2}\right)\\\overset{{\color{red}4}}{\cancel{20}} \left(\frac{2}{\cancel{5}}\right)m+\overset{{\color{red}5}}{\cancel{20}} \left(\frac{3}{\cancel{4}}\right)n &= \overset{{\color{red}10}}{\cancel{20}} \left(\frac{5}{\cancel{2}}\right)\\{\color{red}8}m+{\color{red}15}n &= {\color{red}50}\\8m+15n &= 50

-\frac{2}{3}m+\frac{1}{2}n=\frac{3}{4} The LCM for the denominators is 12.

-{\color{red}12} \left(\frac{2}{3}\right)m+{\color{red}12} \left(\frac{1}{2}\right)n &= {\color{red}12} \left(\frac{3}{4}\right)\\\overset{{\color{red}4}}{-\cancel{12}} \left(\frac{2}{\cancel{3}}\right)m+\overset{{\color{red}6}}{\cancel{12}} \left(\frac{1}{\cancel{2}}\right)n &= \overset{{\color{red}3}}{\cancel{12}}\left(\frac{3}{\cancel{4}}\right)\\{\color{red}-8}m+{\color{red}6}n &= {\color{red}9}\\-8m+6n &= 9

The two equations that need to be solved by substitution are: \begin{Bmatrix}8m+15n=50 \\-8m+6n=9\end{Bmatrix}

Neither of the equations have a variable with a coefficient of 1 nor does one equation have coefficients that are multiples of a given coefficient. Solve the first equation in terms of ‘m’.

& 8m+15n = 50\\& 8m+15n {\color{red}-15n} = 50 {\color{red}-15n}\\& 8m = 50-15n\\& \frac{8m}{{\color{red}8}} = \frac{50}{{\color{red}8}}-\frac{15n}{{\color{red}8}}\\& \frac{\cancel{8}m}{\cancel{8}} = \frac{50}{8}-\frac{15n}{8}\\& m = {\color{red}\frac{25}{4}}-\frac{15}{8}n\\& \boxed{m = \frac{25}{4}-\frac{15}{8}n}

Substitute this value for ‘m’ into the second equation.

-8m+6n &= 9\\-8 \left({\color{red}\frac{25}{4}-\frac{15}{8}n}\right)+6n &= 9

Apply the distributive property and solve the equation.

& {\color{red}-\frac{200}{4}+\frac{120}{8}n}+6n = 9\\& -\frac{\overset{{\color{red}50}}{\cancel{200}}}{\cancel{4}}+\frac{\overset{{\color{red}15}}{\cancel{120}}}{\cancel{8}}n+6n = 9\\& {\color{red}-50+15n}+6n = 9\\& -50 {\color{red}+21n} = 9\\& -50 {\color{red}+50}+21n = 9 {\color{red}+50}\\& 21n = {\color{red}59}\\& \frac{21n}{{\color{red}21}} = \frac{59}{{\color{red}21}}\\& \frac{\cancel{21}n}{\cancel{21}} = \frac{59}{21}\\& \boxed{n = \frac{59}{21}}

Substitute this value into the equation that has been solved in terms of ‘m’ or into one of the original equations or into one of the new equations that resulted from multiplying by the LCM.

Whichever substitution is performed, the same result will occur.

m &= \frac{25}{4}-\frac{15}{8}n\\m &= \frac{25}{4}-\frac{15}{8} \left({\color{red}\frac{59}{21}}\right)\\m &= \frac{25}{4}-{\color{red}\frac{885}{168}}

A common denominator is required to subtract the fractions.

& \overset{ \quad \ {\color{red}42}}{4 \overline{ ) {168}}}\\& \underline{- 16 {\color{blue}\downarrow}}\\& \quad \ \ 8\\& \underline{- \;\;\;8}\\& \quad \ \ 0

Multiply \frac{25}{4} by {\color{red}\frac{42}{42}} :

& m = {\color{red}\frac{42}{42}} \left(\frac{25}{4}\right)-\frac{885}{168}\\& m = {\color{red}\frac{1050}{168}}-\frac{885}{168}\\& m= {\color{red}\frac{165}{168}}\\& m = {\color{red}\frac{55}{56}}\\& \boxed{m = \frac{55}{56}}\\& \boxed{l_1 \cap l_2 @ \left(\frac{55}{56}, \frac{59}{21}\right)}

Summary

In this lesson you have learned that a 2 \times 2 system of linear equations can be solved algebraically by the substitution method. This method involves solving one of the equations in terms of a variable and substituting that expression into the remaining equation. The result will be a linear equation, with one variable, that can be solved. The solution to this equation can be substituted into the solved equation or into the other equation to determine the value of the other variable. The solution is the intersection point of the two equations and it represents the coordinates of the ordered pair.

You have also learned another form of substitution known as the comparison method. This method is useful when both equations are expressed in terms of the same variable or expression. The equations are set equal to each other and solved.

The last exercise in the guided practice showed how involved solving a system of equations by substitution can be. In the next lesson you will learn another method for solving systems of linear equations that will reduce the difficulty of solving such a system.

Problem Set

Solve the following systems of linear equations using the substitution method.

\begin{Bmatrix}2x+y=5 \\3x-4y=2\end{Bmatrix}

\begin{Bmatrix}5x-2y=-4 \\4x+y=-11\end{Bmatrix}

\begin{Bmatrix}3y-x=-10 \\3x+4y=-22\end{Bmatrix}

\begin{Bmatrix}4e+2f=-2 \\2e-3f=1\end{Bmatrix}

\begin{Bmatrix}\frac{1}{4}x+y=-\frac{7}{2} \\\frac{1}{2}x-\frac{1}{4}y=1\end{Bmatrix}

Solve each of the following linear systems using the comparison method

\begin{Bmatrix}x=-4+y \\x=3y-6\end{Bmatrix}

\begin{Bmatrix}3y-2x=-3 \\3x-3y=6\end{Bmatrix}

\begin{Bmatrix}2x=5y-12 \\3x+5y=7\end{Bmatrix}

\begin{Bmatrix}3y=2x-5 \\2x=y+3\end{Bmatrix}

\begin{Bmatrix}\frac{x+y}{3}+\frac{x-y}{2}=\frac{25}{6} \\\frac{x+y-9}{2}=\frac{y-x-6}{3}\end{Bmatrix}

Answers

Solve the following systems...

\begin{Bmatrix}2x+y=5 \\3x-4y=2\end{Bmatrix}

& 2x+y = 5\\& 2x-2x+y = 5-2x\\& \boxed{y = 5-2x}

& 3x-4y = 2\\& 3x-4(5-2x) = 2\\& 3x-20+8x = 2\\& 11x-20 = 2\\& 11x-20+20 = 2+20\\& 11x = 22\\& \frac{11x}{11} = \frac{22}{11}\\& \boxed{x = 2}

& y=5-2x\\& y = 5-2(2)\\& y = 5-4\\& \boxed{y = 1}\\& \boxed{l_1 \cap l_2 @ (2,1)}

\begin{Bmatrix}3y-x=-10 \\3x+4y=-22\end{Bmatrix}

& 3y-x = -10\\& 3y-3y-x = -10-3y\\& -x = -10-3y\\& \frac{-1x}{-1} = \frac{-10}{-1}-\frac{3y}{-1}\\& \frac{\cancel{-1}x}{\cancel{-1}} = \frac{-10}{-1}-\frac{3y}{-1}\\& \boxed{x = 10+3y}

& 3x+4y = -22\\& 3(10+3y)+4y = -22\\& 30+9y+4y = -22\\& 30+13y = -22\\& 30-30+13y = -22-30\\& 13y =-52\\& \frac{13y}{13} = \frac{-52}{13}\\& \frac{\cancel{13}y}{\cancel{13}} = \frac{\overset{-4}{\cancel{-52}}}{\cancel{13}}\\& \boxed{y = -4}

& x=10+3y\\& x = 10+3(-4)\\& x = 10-12\\& \boxed{x = -2}\\& \boxed{l_1 \cap l_2 @ (-2,-4)}

\begin{Bmatrix}\frac{1}{4}x+y =- \frac{7}{2} \\\frac{1}{2} x - \frac{1}{4}y=1\end{Bmatrix}

& \frac{1}{4}x+y = -\frac{7}{2}\\& 4 \left(\frac{1}{4}\right)x+4(y) = -4 \left(\frac{7}{2}\right)\\& \cancel{4} \left(\frac{1}{\cancel{4}}\right)x+4(y) = - \overset{2}{\cancel{4}} \left(\frac{7}{\cancel{2}}\right)\\& \boxed{x+4y = -14}

& \frac{1}{2}x-\frac{1}{4}y = 1\\& 4 \left(\frac{1}{2}\right)x-4 \left(\frac{1}{4}\right)y = 4(1)\\& \overset{2}{\cancel{4}} \left(\frac{1}{\cancel{2}}\right)x- \cancel{4} \left(\frac{1}{\cancel{4}}\right)y = 4(1)\\& \boxed{2x-y = 4}

\begin{Bmatrix}x+4y=-14 \\2x-y=4\end{Bmatrix}

& x+4y = -14\\& x+4y-4y = -14-4y\\& \boxed{x = -14-4y}\\& 2x-y = 4\\& 2(-14-4y)-y = 4\\& -28-8y-y = 4\\& -28-9y = 4\\& -28+28-9y = 4+28\\& -9y = 32\\& \frac{-9y}{-9} = \frac{32}{-9}\\& \frac{\cancel{-9}y}{\cancel{-9}} = \frac{32}{-9}\\& \boxed{y = -\frac{32}{9}}

& x = -14-4y\\& x = -14-4 \left(-\frac{32}{9}\right)\\& x = -14+\frac{128}{9}\\& x = -\frac{126}{9}+\frac{128}{9}\\& \boxed{x = \frac{2}{9}}\\& \boxed{l_1 \cap l_2 @ \left(\frac{2}{9},-\frac{32}{9}\right)}

Solve each of the following...

& \begin{Bmatrix}x=-4+y\\x=3y-6\end{Bmatrix}

& -4+y = 3y-6\\& -4+4+y = 3y-6+4\\& y = 3y-2\\& y-3y = 3y-3y-2\\& -2y = -2\\& \frac{-2y}{-2} = \frac{-2}{-2}\\& \boxed{y = 1}

& x = -4+y\\& x = -4+1\\& \boxed{x = -3}\\& \boxed{l_1 \cap l_2 @ (-3,1)}

& \begin{Bmatrix}2x=5y-12\\3x+5y=7\end{Bmatrix}

& 2x = 5y-12\\& 2x+12 = 5y-12+12\\& 2x+12 = 5y\\& \boxed{5y = 2x+12}

& 3x+5y = 7\\& 3x-3x+5y = 7-3x\\& \boxed{5y = 7-3x}

& \begin{Bmatrix}5y=2x+12\\5y=7-3x\end{Bmatrix}

& 2x+12 = 7-3x\\& 2x+12+3x = 7-3x+3x\\& 5x+12 = 7\\& 5x+12-12 = 7-12\\& 5x = -5\\& \frac{5x}{5} = \frac{-5}{5}\\& \frac{\cancel{5}x}{\cancel{5}} = \frac{\overset{-1}{\cancel{-5}}}{\cancel{5}}\\& \boxed{x = -1}

& 5y = 7-3x\\& 5y = 7-3(-1)\\& 5y = 7+3\\& 5y = 10\\& \frac{5y}{5} = \frac{10}{5}\\& \frac{\cancel{5}y}{\cancel{5}} = \frac{\overset{2}{\cancel{10}}}{\cancel{5}}\\& \boxed{y = 2}\\& \boxed{l_1 \cap l_2 @ (-1,2)}

& \begin{Bmatrix}\frac{x+y}{3}+\frac{x-y}{2} =\frac{25}{6}\\\frac{x+y-9}{2} =\frac{y-x-6}{3}\end{Bmatrix}

& \frac{x+y}{3}+\frac{x-y}{2} = \frac{25}{6}\\& 6 \left(\frac{x+y}{3}\right)+6 \left(\frac{x-y}{2}\right) = 6 \left(\frac{25}{6}\right)\\& \overset{2}{\cancel{6}} \left(\frac{x+y}{\cancel{3}}\right)+\overset{3}{\cancel{6}} \left(\frac{x-y}{\cancel{2}}\right) = \cancel{6} \left(\frac{25}{\cancel{6}}\right)\\& 2(x+y)+3(x-y) = 25\\& 2x+2y+3x-3y = 25\\& \boxed{5x-y = 25}

& \frac{x+y-9}{2} = \frac{y-x-6}{3}\\& 6 \left(\frac{x+y-9}{2}\right) = 6 \left(\frac{y-x-6}{3}\right)\\& \overset{3}{\cancel{6}} \left(\frac{x+y-9}{\cancel{2}}\right) = \overset{2}{\cancel{6}} \left(\frac{y-x-6}{\cancel{3}}\right)\\& 3(x+y-9) = 2(y-x-6)\\& 3x+3y-27 = 2y-2x-12\\& 3x+3y-27+27 = 2y-2x-12+27\\& 3x+3y = 2y-2x+15\\& 3x+3y+2x = 2y-2x+2x+15\\& 5x+3y = 2y+15\\& 5x+3y-2y = 2y-2y+15\\& \boxed{5x+y = 15}

& \begin{Bmatrix}5x-y=25\\5x+y=15\end{Bmatrix}

& 5x-y = 25 && 5x+y=15\\& 5x-y+y = 25+y && 5x+y-y=15-y\\& \boxed{5x = 25+y} && \boxed{5x=15-y}\\& 25+y = 15-y\\& 25-25+y = 15-y-25\\& y = -y-10\\& y+y = -y+y-10\\& 2y = -10\\& \frac{2y}{2} = \frac{-10}{2}\\& \frac{\cancel{2}y}{\cancel{2}} = \frac{\overset{-5}{\cancel{-10}}}{\cancel{2}}\\& \boxed{y =-5}

& 5x = 25+y\\& 5x = 25+(-5)\\& 5x = 20\\& \frac{5x}{5} = \frac{20}{5}\\& \frac{\cancel{5}x}{\cancel{5}} = \frac{\overset{4}{\cancel{20}}}{\cancel{5}}\\& \boxed{x = 4}\\& \boxed{l_1 \cap l_2 @ (4,-5)}

Solving Systems of Linear Equations by Elimination

Introduction

In this lesson you will learn to solve systems of linear equations algebraically. When technology is not available, the solution to a consistent and independent system can be determined by using algebra. This is necessary if the solution is not integers.

The algebraic method that you will learn in this lesson is called elimination. This method is also referred to as the addition and subtraction method. This method involves obtaining an equivalent system of equations such that, when two of the equations are added or subtracted, one of the variables is eliminated. A variable can only be eliminated when its coefficient is zero. This method applies the zero principle.

Objectives

The lesson objectives for Solving Systems of Linear Equations Algebraically are:

  • Determining the solution by elimination

Introduction

In this lesson you will learn to solve a linear equation that has two variables by substitution. A 2 \times 2 system of linear equations consists of two equations with two variables. Elimination involves multiplying one or both of the equations to create equations in which the same variable in each equation has the same numerical coefficient but opposite signs. When the equations are combined, that variable will have a coefficient of zero – in other words, that variable will be eliminated. This will result in a linear equation with one variable. The solution to this equation will then be substituted into one of original equations – again resulting in a linear equation with one variable. The solution to this equation will be the value for the variable that was eliminated. The two values together represent the x and y-values of the point of intersection for the system of equations.

Watch This

Khan Academy Solving Systems by Elimination

Khan Academy Solving Systems by Elimination 2

Guidance

When you solve a system of consistent and independent equations by elimination, a single ordered pair is the solution. The ordered pair satisfies both equations and the point is the intersection of the graphs of the linear equations. The coordinates of this point of intersection is not always integers. Therefore, some method has to be used to determine the values of the coordinates. The method that will be presented here is called the elimination method.

Solve by substitution:

\begin{Bmatrix}2x+3y=5\\3x-3y=10\end{Bmatrix}

Both equations have a term that is 3y. In the first equation the coefficient of ‘y’ is a positive three and in the second equation the coefficient of ‘y’ is a negative three. If the two equations were added, the ‘y’ variable would be eliminated.

& 2x+3y=5\\& \underline{3x-3y=10}\\& 5x+{\color{red}0y}=15

The coefficient of ‘y’ is zero. The coefficient of a variable multiplies that variable. Then

0y={\color{red}0}(y)={\color{red}0}

Therefore the variable has been eliminated. This can be shown in the addition of the two equations as:

& 2x+{\cancel{3y}}=5 && && 2x+3y=5\\& \underline{3x-\cancel{3y}=10} && && \underline{3x-3y=10}\\& \qquad \ 5x=15 && \text{And not} && 5x+ {\color{red}0y}=15

The resulting equation now has one variable. Solve this equation:

& 5x = 15\\& \frac{5x}{{\color{red}5}} = \frac{15}{{\color{red}5}}\\& \frac{\cancel{5}x}{\cancel{5}} = \frac{\overset{{\color{red}3}}{\cancel{15}}}{\cancel{5}}\\& \boxed{x = 3}

The value of ‘x’ is 3. This value can now be substituted into one of the original equations to determine the value of ‘y’. Remember ‘y’ is the variable that was eliminated from the system of linear equations.

& 2x+3y = 5\\& 2({\color{red}3})+3y = 5 && \text{Substitute in the value for} \ `x'.\\& {\color{red}6}+3y = 5 && \text{Multiply the value of} \ `x' \ \text{by the coefficient} \ (2).\\& 6 {\color{red}-6}+3y = 5 {\color{red}-6} && \text{Isolate the variable} \ `y'.\\& 3y = {\color{red}-1} && \text{Solve the equation.}\\& \frac{3y}{{\color{red}3}} = \frac{-1}{{\color{red}3}}\\& \frac{\cancel{3}y}{\cancel{3}} = {\color{red}-\frac{1}{3}}\\& \boxed{y = -\frac{1}{3}}

The solution to the system of linear equations is x=3 and y=-\frac{1}{3}. This solution means \boxed{l_1 \cap l_2 @ \left(3,-\frac{1}{3}\right)}

Example A

Solve the following system of linear equations by elimination:

\begin{Bmatrix}3y=2x-5\\2x=y+3\end{Bmatrix}

To begin, set up the equations such that they are in the format \begin{Bmatrix}a_1 {\color{red}x}+b_1 {\color{blue}y}=c_1\\a_2 {\color{red}x}+b_2 {\color{blue}y}=c_2\end{Bmatrix}

3y &= 2x-5 && 2x=y+3\\{\color{red}-2x}+3y &= 2x {\color{red}-2x}-5 && 2x {\color{red}-y} = y {\color{red}-y}+3\\-2x+3y &= -5 && 2x-y=3

Solve the formatted system of equations:

\begin{Bmatrix}-2x+3y=-5\\2x-y=3\end{Bmatrix}

Both equations have a term that is 2x. In the first equation the coefficient of ‘x’ is a negative two and in the second equation the coefficient of ‘x’ is a positive two. If the two equations were added, the ‘x’ variable would be eliminated.

& -\cancel{2x}+3y=-5\\& \quad \underline{\cancel{2x}- \; y=+3 \;}\\& \qquad \quad {\color{red}2y=-2} \quad \text{Eliminate the variable} \ `x'.\\& 2y = -2 \quad \text{Solve the equation.}\\& \frac{2y}{{\color{red}2}} = \frac{-2}{{\color{red}2}}\\& \frac{\cancel{2}y}{\cancel{2}} = \frac{\overset{{\color{red}-1}}{\cancel{-2}}}{\cancel{2}}\\& \boxed{y = -1}

The value of ‘y’ is -1. This value can now be substituted into one of the original equations to determine the value of ‘x’. Remember ‘x’ is the variable that was eliminated from the system of linear equations.

& 2x-y = 3\\& 2x-({\color{red}-1}) = 5 && \text{Substitute in the value for} \ `y'.\\& 2x {\color{red}+1} = 5 && \text{Multiply the value of} \ `x' \ \text{by the coefficient} \ (-1).\\& 2x+1 {\color{red}-1} = 3 {\color{red}-1} && \text{Isolate the variable} \ `x'.\\& 2x = {\color{red}2} && \text{Solve the equation.}\\& \frac{2x}{{\color{red}2}} = \frac{2}{{\color{red}2}}\\& \frac{\cancel{2}x}{\cancel{2}} = \frac{\overset{{\color{red}1}}{\cancel{2}}}{\cancel{2}}\\& \boxed{x = 1}\\& \boxed{l_1 \cap l_2 @ (1,-1)}

Example B

Solve the following system of linear equations by elimination:

\begin{Bmatrix}2x-3y=13\\3x+4y=-6\end{Bmatrix}

The coefficients of ‘x’ are 2 and 3. The coefficients of ‘y’ are -3 and 4. To eliminate a variable the coefficients must be the same number but with opposite signs. This can be accomplished by multiplying one or both of the equations.

The first step is to choose a variable to eliminate. If the choice is ‘x’, the least common multiple of 2 and 3 is 6. This means that the equations must be multiplied by 3 and 2 respectively. This will create the variable ‘x’ in both equations having a coefficient of 6. However, when the equations are added, the variable will not have a coefficient of zero. One of the multipliers must be a negative number so that one of the coefficients of ‘x’ will be a negative 6. When this is done, the coefficients of ‘x’ will be +6 and -6. The variable will then be eliminated when the equations are added.

Multiply the first equation by negative three.

& {\color{red}-3}(2x-3y = 13)\\& {\color{red}-6x+9y=-39}

Multiply the second equation by positive two.

& {\color{red}2}(3x+4y=-6)\\& {\color{red}6x+8y=-12}

Add the two equations.

& -\cancel{6x}+9y=-39\\& \ \ \underline{\cancel{6x}+8y=-12}\\& \qquad \ \ {\color{red}17y=-51} \qquad \text{Solve the equation.}\\& 17y =- 51\\& \frac{17y}{{\color{red}17}} = \frac{-51}{{\color{red}17}}\\& \frac{\cancel{17}y}{\cancel{17}} = \frac{\overset{{\color{red}-3}}{\cancel{-51}}}{\cancel{17}}\\& \boxed{y = -3}

Substitute the value for ‘y into one of the original equations.

& 2x-3y = 13\\& 2x-3({\color{red}-3}) = 13 && \text{Substitute in the value for} \ `y'.\\& 2x {\color{red}+9}=13 && \text{Multiply the value of} \ `y' \ \text{by the coefficient} \ (-3).\\& 2x+9 {\color{red}-9} = 13 {\color{red}-9} && \text{Isolate the variable} \ `x'.\\& 2x = {\color{red}4} && \text{Solve the equation.}\\& \frac{2x}{{\color{red}2}} = \frac{4}{{\color{red}2}}\\& \frac{\cancel{2}x}{\cancel{2}} = \frac{\overset{{\color{red}2}}{\cancel{4}}}{\cancel{2}}\\& \boxed{x = 2}\\& \boxed{l_1 \cap l_2 @ (2,-3)}

Example C

Solve the following system of linear equations by elimination:

\begin{Bmatrix}\frac{3}{4}x+\frac{5}{4}y = 4\\\frac{1}{2}x+\frac{1}{3}y = \frac{5}{3}\end{Bmatrix}

Begin by multiplying each equation by the LCD to create two equations with integers as the coefficients of the variables.

\frac{3}{4}x+\frac{5}{4}y &= 4 && \frac{1}{2}x+\frac{1}{3}y=\frac{5}{3}\\{\color{red}4} \left(\frac{3}{4}\right)x+ {\color{red}4} \left(\frac{5}{4}\right)y &= {\color{red}4} (4) && {\color{red}6} \left(\frac{1}{2}\right)x+ {\color{red}6} \left(\frac{1}{3}\right)y= {\color{red}6} \left(\frac{5}{3}\right)\\\cancel{4} \left(\frac{3}{\cancel{4}}\right)x+\cancel{4} \left(\frac{5}{\cancel{4}}\right)y &= 4(4) && \overset{{\color{red}3}}{\cancel{6}} \left(\frac{1}{\cancel{2}} \right)x+\overset{{\color{red}2}}{\cancel{6}} \left(\frac{1}{\cancel{3}} \right)y=\overset{{\color{red}2}}{\cancel{6}}\left(\frac{5}{\cancel{3}}\right)\\{\color{red}3}x+{\color{red}5}y &= {\color{red}16} && {\color{red}3}x+{\color{red}2}y={\color{red}10}

Solve the following system of equations by elimination:

\begin{Bmatrix}3x+5y=16\\3x+2y=10\end{Bmatrix}

The coefficients of the ‘x’ variable are the same – positive three. To change one of them to a negative three, multiply one of the equations by a negative one.

& {\color{red}-1} (3x+5y=16)\\& {\color{red}-3x-5y=-16}

The two equations can now be added.

& - \cancel{3x}-5y=-16\\& \ \ \underline{\cancel{3x}+2y= \;\;\; 10}\\& \qquad {\color{red}-3y= \ \ -6} \qquad \text{Solve the equation.}\\& -3y = -6\\& \frac{-3y}{{\color{red}-3}} = \frac{-6}{{\color{red}-3}}\\& \frac{\cancel{-3}y}{{\cancel{-3}}} = \frac{\overset{{\color{red}2}}{\cancel{-6}}}{\cancel{-3}}\\& \boxed{y = 2}

Substitute the value for ‘y’ into one of the original equations.

& \frac{3}{4}x+\frac{5}{4}y = 4\\& \frac{3}{4}x+\frac{5}{4}({\color{red}2}) = 4 && \text{Substitute in the value for} \ `y'.\\& \frac{3}{4}x+\frac{{\color{red}10}}{4}=4 && \text{Multiply the value of} \ `y' \ \text{by the coefficient} \ \left(\frac{5}{4}\right).\\& \frac{3}{4}x+\frac{10}{4}- {\color{red}\frac{10}{4}} = 4-{\color{red}\frac{10}{4}} && \text{Isolate the variable} \ `x'.\\& \frac{3}{4}x = {\color{red}\frac{16}{4}}- {\color{red}\frac{10}{4}}\\& \frac{3}{4}x = {\color{red}\frac{6}{4}} && \text{Multiply both sides by} \ 4.\\& {\color{red}4} \left(\frac{3}{4}x\right) = {\color{red}4} \left(\frac{6}{4}\right)\\& {\cancel{4}} \left(\frac{3}{\cancel{4}}x\right) = \cancel{4} \left(\frac{6}{\cancel{4}}\right)\\& 3x = 6 && \text{Solve the equation.}\\& \frac{3x}{{\color{red}3}} = \frac{6}{{\color{red}3}}\\& \frac{\cancel{3}x}{\cancel{3}} = \frac{\overset{{\color{red}2}}{\cancel{6}}}{\cancel{3}}\\& \boxed{x = 2}\\& \boxed{l_1 \cap l_2 @ (2,2)}

Vocabulary

Elimination Method
The elimination method is a method used for solving a system of linear equations algebraically. This method involves obtaining an equivalent system of equations such that, when two of the equations are added or subtracted, one of the variables is eliminated.

Guided Practice

1. Solve the following system of linear equations by comparison:

\begin{Bmatrix}4x-15y=5\\6x-5y=4\end{Bmatrix}

2. Solve the following system of linear equations by substitution:

\begin{Bmatrix}3x=7y+41\\5x=3y+51\end{Bmatrix}.

3. Solve the following system of linear equations by elimination:

\begin{Bmatrix}\frac{2}{5}m+\frac{3}{4}n=\frac{5}{2}\\-\frac{2}{3}m+\frac{1}{2}n=\frac{3}{4}\end{Bmatrix}

Answers

1. \begin{Bmatrix}4x-15y=5\\6x-5y=4\end{Bmatrix}

Multiply the second equation by (-3) to eliminate the variable ‘y’.

& {\color{red}-3} (6x-5y=4)\\ & {\color{red}-18x+15y=-12}\\& -18x+15y=-12

Add the equations:

& \quad \ 4x - \cancel{15y}=5\\& \underline{-18x + \cancel{15y}=-12}\\& {\color{red}-14x=-7}

Solve the equation:

& -14x = -7\\& \frac{-14x}{{\color{red}-14}} = \frac{-7}{{\color{red}-14}}\\& \frac{\cancel{-14}x}{\cancel{-14}} = \frac{-7}{-14}\\& \boxed{x = \frac{1}{2}}

Substitute this value for ‘x’ into one of the original equations.

& 4x-15y = 5\\& 4 \left({\color{red}\frac{1}{2}}\right)-15y = 5 && \text{Substitute in the value for} \ `x'.\\& {\color{red}2}-15y = 5 && \text{Multiply the value of} \ `x' \ \text{by the coefficient} \ (4).\\& 2 {\color{red}-2}-15y = 5 {\color{red}-2} && \text{Isolate the variable} \ `y'.\\& -15y = 3 && \text{Solve the equation.}\\& \frac{-15y}{{\color{red}-15}} = \frac{3}{{\color{red}-15}}\\& \frac{\cancel{-15}y}{\cancel{-15}} = \frac{3}{-15}\\& \boxed{y = -\frac{1}{5}}\\& \boxed{l_1 \cap l_2 @ \left(\frac{1}{2}, -\frac{1}{5}\right)}

2. \begin{Bmatrix}3x=7y+41\\5x=3y+51\end{Bmatrix}

Arrange the equations so that they are of the form \begin{Bmatrix}a_1 {\color{red}x}+b_1 {\color{blue}y}=c_1\\a_2 {\color{red}x}+b_2 {\color{blue}y}=c_2\end{Bmatrix}.

3x &= 7y+41 && 5x=3y+51\\3x {\color{red}-7y} &= 7y {\color{red}-7y}+41 && 5x {\color{red}-3y}=3y {\color{red}-3y}+51\\3x-7y &=41 && 5x-3y=51

Multiply the first equation by (-5) and the second equation by (3).

& {\color{red}-5}(3x-7y=41) && {\color{red}3} (5x-3y=51)\\& {\color{red}-15x+35y=-205} && {\color{red}15x-9y=153}\\& -15x+35y=-205 && 15x-9y=153

Add the equations to eliminate ‘x’.

& - \cancel{15x}+35y=-205\\& \quad \underline{\;\;\; \cancel{15x}-9y=153\;\;\;}\\& \qquad \qquad {\color{red}26y=-52}

Solve the equation:

& 26y = 52\\& \frac{26y}{{\color{red}26}} = \frac{-52}{{\color{red}26}}\\& \frac{\cancel{26}y}{\cancel{26}} = \frac{\overset{{\color{red}-2}}{\cancel{-52}}}{\cancel{26}}\\& \boxed{y =-2}\\& 5x-3y = 51\\& 5x-3({\color{red}-2}) = 51 && \text{Substitute in the value for} \ `y'.\\& 5x {\color{red}+6}=51 && \text{Multiply the value of} \ `y' \ \text{by the coefficient} \ (-3).\\& 5x+6 {\color{red}-6} = 51 {\color{red}-6} && \text{Isolate the variable} \ `x'.\\& 5x = {\color{red}45} && \text{Solve the equation.}\\& \frac{5x}{{\color{red}5}} = \frac{45}{{\color{red}5}}\\& \frac{\cancel{5}x}{\cancel{5}} = \frac{\overset{{\color{red}9}}{\cancel{45}}}{\cancel{5}}\\& \boxed{x = 9}\\& \boxed{l_1 \cap l_2 @ (9,-2)}

3. \begin{Bmatrix}\frac{2}{5}m+\frac{3}{4}n=\frac{5}{2}\\-\frac{2}{3}m+\frac{1}{2}n=\frac{3}{4}\end{Bmatrix}

This problem was done in the lesson solving systems of linear equations by substitution. Here it will be done by elimination. The difference in the two methods will be visible.

Begin by multiplying each equation by the LCM of the denominators to simplify the system.

\frac{2}{5}m+\frac{3}{4}n=\frac{5}{2} The LCM for the denominators is 20.

& {\color{red}20} \left(\frac{2}{5}\right)m+{\color{red}20} \left(\frac{3}{4}\right)n = {\color{red}20} \left(\frac{5}{2}\right)\\& \overset{{\color{red}4}}{\cancel{20}} \left(\frac{2}{\cancel{5}}\right)m+\overset{{\color{red}5}}{\cancel{20}} \left(\frac{3}{\cancel{4}} \right)n = \overset{{\color{red}10}}{\cancel{20}} \left(\frac{5}{\cancel{2}}\right)\\& {\color{red}8}m+{\color{red}15}n = {\color{red}50}\\& \boxed{8m+15n = 50}\\& -\frac{2}{3}m+\frac{1}{2}n = \frac{3}{4} && \text{The LCM for the denominators is} \ 12.\\& -{\color{red}12} \left(\frac{2}{3}\right)m+{\color{red}12} \left(\frac{1}{2}\right)n = {\color{red}12} \left(\frac{3}{4}\right)\\& -\overset{{\color{red}4}}{\cancel{12}} \left(\frac{2}{\cancel{3}}\right)m+\overset{{\color{red}6}}{\cancel{12}} \left(\frac{1}{\cancel{2}}\right)n = \overset{{\color{red}3}}{\cancel{12}} \left(\frac{3}{\cancel{4}}\right)\\& {\color{red}-8}m+{\color{red}6}n = {\color{red}9}\\& \boxed{-8m+6n = 9}

The two equations that need to be solved are: \begin{Bmatrix}8m+15n=50\\-8m+6n=9\end{Bmatrix}

The equations will be solved by using the elimination method. The variable ‘m’ has the same numerical coefficient with opposite signs. The variable will be eliminated when the equations are added.

& \cancel{8m}+15n = 50\\& \underline{-\cancel{8m}+6n=9}\\& \qquad {\color{red}21n=59}

Solve the equation:

& 21n = 59\\& \frac{21n}{{\color{red}21}} = \frac{59}{{\color{red}21}}\\& \frac{\cancel{21}n}{\cancel{21}} = \frac{59}{21}\\& \boxed{n = \frac{59}{21}}\\& \frac{2}{5}m+\frac{3}{4}n = \frac{5}{2}\\& \frac{2}{5}m+\frac{3}{4} \left({\color{red}\frac{59}{21}}\right) = \frac{5}{2} && \text{Substitute in the value for} \ `n'.\\& \frac{2}{5}m+{\color{red}\frac{177}{84}} = \frac{5}{2} && \text{Multiply the value of} \ `y' \ \text{by the coefficient} \left(\frac{59}{21}\right).\\& \frac{2}{5}m+\frac{177}{84}-{\color{red}\frac{177}{84}} = \frac{5}{2}- {\color{red}\frac{177}{84}} && \text{Isolate the variable} \ `x'.\\& \frac{2}{5}m = {\color{red}\frac{210}{84}}-{\color{red}\frac{177}{84}}\\& \frac{2}{5}m = {\color{red}\frac{33}{84}} && \text{Solve the equation.}\\& {\color{red}420} \left(\frac{2}{5}\right)x = {\color{red}420} \left(\frac{33}{84}\right)\\& \overset{{\color{red}84}}{\cancel{420}} \left(\frac{2}{\cancel{5}}\right)x = \overset{{\color{red}5}}{\cancel{420}} \left(\frac{33}{\cancel{84}}\right)\\& {\color{red}168}x = {\color{red}165}\\& \frac{168x}{{\color{red}168}} = \frac{165}{{\color{red}168}}\\& \frac{\cancel{168}x}{{\cancel{168}}} = \frac{165}{168}\\& \boxed{x = \frac{55}{56}}\\& \boxed{l_1 \cap l_2 @ \left(\frac{55}{56}, \frac{59}{21}\right)}

Summary

In this lesson you have learned that a 2 \times 2 system of linear equations can be solved algebraically by the elimination method. This method involves obtaining an equivalent system of equations such that, when two of the equations are added or subtracted, one of the variables is eliminated. The solution is the intersection point of the two equations and it represents the coordinates of the ordered pair.

The last exercise in the guided practice of the previous lesson showed how involved solving a system of equations by substitution can become. In this lesson you have done the same problem using the elimination method. This method was not as involved as was solving the problem using the substitution method.

Problem Set

Solve the following systems of linear equations using the elimination method.

\begin{Bmatrix}16x-y-181=0\\19x-y=214\end{Bmatrix}

\begin{Bmatrix}3x+2y+9=0\\4x=3y+5\end{Bmatrix}

\begin{Bmatrix}x=7y+38\\14y=-x-46\end{Bmatrix}

\begin{Bmatrix}2x+9y=-1\\4x+y=15\end{Bmatrix}

\begin{Bmatrix}x-\frac{3}{5}y=\frac{26}{5}\\4y=61-7x\end{Bmatrix}

Solve each of the following linear systems using the elimination method.

\begin{Bmatrix}3x-5y=12\\2x+10y=4\end{Bmatrix}

\begin{Bmatrix}3x+2y+9=0\\4x=3y+5\end{Bmatrix}

\begin{Bmatrix}x=69+6y\\3x=4y-45\end{Bmatrix}

\begin{Bmatrix}3(x-1)-4(y+2)=-5\\4(x+5)-(y-1)=16\end{Bmatrix}

\begin{Bmatrix}\frac{3}{4}x-\frac{2}{5}y=2\\\frac{1}{7}x+\frac{3}{2}y=\frac{113}{7}\end{Bmatrix}

Answers

Solve the following systems...

& \begin{Bmatrix}16x-y-181=0\\19x-y=214\end{Bmatrix}\\& 16x-y-181= 0 && 16x-y=181 && -1(16x-y=181)\\& 16x-y-181+181=0+181 && 19x-y=214 \quad \Rightarrow && 19x-y=214\\& 16x-y=181\\& && -16x+y=-181 && 16x-y=181\\& && \underline{19x-y=214\;\;\;\;\;\;} && 16(11)-y=181\\& && 3x = 33 && 176-y=181\\& && \frac{3}{3}x=\frac{33}{3} && 176-176-y=181-176\\& && x=11 && \frac{-1}{-1}y=\frac{5}{-1}\\& && && y=-5

The solution is (11, -5).

& \begin{Bmatrix}x=7y+38\\14y=-x-46\end{Bmatrix}\\& x=7y+38 && 14y=-x-46\\& x-7y=7y-7y+38 && 14y+x=-x+x-46\\& x-7y=38 && x+14y=-46\\\\& x-7y=38 && 2(x-7y=38)\\& x+14y=-46 \quad \Rightarrow && x+14y=-46 \quad \Rightarrow

& && 2x-14y=76 && x=7y+38\\& && \underline{x+14y=-46} && 10=7y+38\\& \Rightarrow && 3x=40 && 10-38=7y+38-38\\& && \frac{3}{3}x=\frac{30}{3} && -28=7y\\& && x=10 && \frac{-28}{7}=\frac{7}{7}y\\& && && -4=y

The solution is (10, -4).

& \begin{Bmatrix}x-\frac{3}{5}y=\frac{26}{5}\\4y=61-7x\end{Bmatrix}\\\\& x-\frac{3}{5}y=\frac{26}{5} && 4y=61-7x\\& 5(x)-5 \left(\frac{3}{5}y\right)=5 \left(\frac{26}{5}\right) && 4y+7x=61-7x+7x\\& 5x-3y=26 && 7x+4y=61\\\\& 5x-3y=26 && 4(5x-3y=26) && 20x-12y=104\\& 7x+4y=61 \quad \Rightarrow && 3(7x+4y=61) \quad \Rightarrow && \underline{21x+12y=183}\\& && && 41x=287\\& && && \frac{41}{41}x=\frac{287}{41}\\& && && x=7

& && 4y=61-7x\\& && 4y=61-7(7)\\& && 4y=61-49\\& \Rightarrow && 4y=12\\& && \frac{4}{4}y=\frac{12}{4}\\& && y=3

The solution is (7, 3)

Solve each of the following...

& \begin{Bmatrix}3x-5y=12\\2x+10y=4\end{Bmatrix}\\\\& 3x-5y=12 \Rightarrow 2(3x-5y=12) \Rightarrow 6x-10y=24\\& 2x+10y=4 \Rightarrow 2x+10y=4 \Rightarrow 2x+10y=4

& 6x-10y=24 && 2x+10y=4\\& \underline{2x+10y=4} && 2(3.5)+10y=4\\& 8x=28 \qquad \quad \Rightarrow && 7+10y=4\\& \frac{8x}{8}=\frac{28}{8} && 7-7+10y=4-7\\& x = 3.5 && 10y=-3\\& && \frac{10y}{10}=\frac{-3}{10}\\& && y=-0.3

The solution is (3.5,-0. 3)

& \begin{Bmatrix}x=69+6y\\3x=4y-45\end{Bmatrix}\\\\& x=69+6y && 3x=4y-45\\& x-6y=69+6y-6y && 3x-4y=-45-4y\\& x-6y=69 && 3x-4y=-45\\\\& x-6y=69 && -3(x-6y=69) && -3x+18y=-207 && x=69+6y\\& 3x-4y=-45 \quad \Rightarrow && 3x-4y=-45 \quad \Rightarrow && \underline{3x-4y=-45\;\;\;\;\;\;\;\;} \qquad \Rightarrow && x=69+6(-18)\\& && && 14y=-252 && x=69-108\\& && && \frac{14}{14}y=\frac{-252}{14} && x=-39\\& && && y=-18

The solution is (-39,-18)

& \begin{Bmatrix}\frac{3}{4}x-\frac{2}{5}y=2\\\frac{1}{7}x+\frac{3}{2}y=\frac{113}{7}\end{Bmatrix}

& && 20 \left(\frac{3}{4}\right)x-20 \left(\frac{2}{5}\right)y=20(2) && 15x-8y=40\\& \Rightarrow && 14 \left(\frac{1}{7}\right)x+14 \left(\frac{3}{2}\right)y=14 \left(\frac{113}{7}\right) \quad \Rightarrow && 2x+21y=226

& && -2(15x-8y=40) && -30x+16y=-80 && \frac{3}{4}x-\frac{2}{5}(10)=2\\& \Rightarrow && 15(2x+21y=226) && \underline{30x+315y=3390\;\;} && \frac{3}{4}x-4=2\\& && && 331y=3310 && \frac{3}{4}x-4+4=2+4\\& && && \frac{331}{331}y=\frac{3310}{331} \qquad \quad \Rightarrow && \frac{3}{4}x=6\\& && && y=10 && (4) \frac{3}{4}x=(4)6\\& && && && 3x=24\\& && && && \frac{3}{3}x=\frac{24}{3}\\& && && && x=8

The solution is (8, 10)

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Jan 16, 2013

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Mar 10, 2014
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CK.MAT.ENG.SE.1.Algebra-I---Honors.5.2

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