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Linear Systems with Multiplication

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Elimination Method for Systems of Equations

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 are not always integers. Therefore, some method has to be used to determine the values of the coordinates. How can you algebraically solve the system of equations below without first rewriting the equations?

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

Watch This

Khan Academy Solving Systems by Elimination

Khan Academy Solving Systems by Elimination 2

Guidance

A 2 \times 2 system of linear equations can be solved algebraically by the elimination method. To use this method you must write 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. This method is demonstrated in the examples.

Example A

Solve the following system of linear equations by elimination:

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

Solution: To begin, set up the equations so 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 are added, the x variable is 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)}

This means "Line 1 intersects Line 2 at the point (1, –1)".

Example B

Solve the following system of linear equations by elimination:

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

Solution: 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. 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}

Solution: 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}

Now 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)}

Concept Problem Revisited

Solve by elimination:

\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 are added, the ‘ y ’ variable is eliminated.

& 2x+3y=5\\& \underline{3x-3y=10}\\& 5x=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 ’.

& 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)}

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 elimination:

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

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

\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}

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)}

Practice

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

  1. .

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

  1. .

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

  1. .

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

  1. .

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

  1. .

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

  1. .

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

  1. .

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

  1. .

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

  1. .

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

  1. .

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

  1. .

\begin{Bmatrix}3x+5y=17\\2x+3y=11\end{Bmatrix}

  1. .

\begin{Bmatrix}3x-5y=-29\\2x-8y=-42\end{Bmatrix}

  1. .

\begin{Bmatrix}7x-8y=-26\\5x-12y=-45\end{Bmatrix}

  1. .

\begin{Bmatrix}6x+5y=5.1\\4x-2y=-1.8\end{Bmatrix}

  1. When does it make sense to use the elimination method to solve a system of equations?

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