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Determining the Type of Linear System

Identify consistent, inconsistent, and dependent systems

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Recognizing Linear Systems
License: CC BY-NC 3.0

Michelle has filled her piggy bank with nickels and quarters. When she empties her bank she has a total of 135 coins. Michelle counts her money and is pleased to know she has a total of $18.75. How many of each coin did Michelle have in her bank?

In this concept, you will learn to recognize linear systems of equations.

Linear System

A linear equation with two variables can be written in standard form such that \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are coefficients of the variables ‘\begin{align*}x\end{align*}’ and ‘\begin{align*}y\end{align*}’ and \begin{align*}C\end{align*} is a constant. The equation is linear since the variables ‘\begin{align*}x\end{align*}’ and ‘\begin{align*}y\end{align*}’ are only to the first power.

\begin{align*}Ax+By=C\end{align*}

A linear system of two equations with two variables is any system that can be written in the form:

\begin{align*}\begin{array}{rcl} A_{1}x+B_{1}y &=& C_{1} \\ A_{2}x+B_{2}y &=& C_{2} \end{array} \end{align*}

The variables ‘\begin{align*}x\end{align*}’ and ‘\begin{align*}y\end{align*}’ must be only in the numerator and there can be no products variables in the equations. An example of a linear system of equations with two variables is shown below:

\begin{align*}\begin{array}{rcl} x+y &=& 12 \\ x-y &=& 4 \end{array}\end{align*}

Remember to solve an equation with two variables means to find an ordered pair \begin{align*}(x,y)\end{align*} that will make the statement of equality true.

Let’s look at the first equation.

\begin{align*}x+y=12\end{align*}

The solution for the equation are any two numbers that have a sum of 12. There are many solutions for this equation. Some examples are \begin{align*}(12,0),(15,-3),(9,3),(25,-13),(8,4)\end{align*} but there are an infinite number of ordered pairs that will make the statement true.

Let’s look at the second equation.

The solution for the equation are any two numbers that have a difference of 4. There are also many solutions for this equation. Some examples are \begin{align*}(12,8),(15,11),(9,5),(8,4),(2,-2)\end{align*} but there are an infinite number of ordered pairs that will make the statement true.

The solution to a system of linear equations is a value for ‘\begin{align*}x\end{align*}’ and a value for ‘\begin{align*}y\end{align*}’ that, when substituted into the equations, makes both equations a true statement of equality.

Let’s test some of the ordered pairs from above into the equations.

Substitute the ordered pairs into the equation to see if it will make the statement true. 

\begin{align*}(9, 3)\end{align*}

First, begin with the first equation.

\begin{align*}x+y=12\end{align*}

Next, substitute \begin{align*}x=9\end{align*} and \begin{align*}y=3\end{align*} into the equation.

\begin{align*}\begin{array}{rcl} x+y &=& 12 \\ 9+3 &=& 12 \end{array}\end{align*}

Then, simplify the left side of the equation.

\begin{align*}\begin{array}{rcl} 9+3 &=& 12 \\ 12 &=& 12 \end{array}\end{align*}

Both sides of the equation are equal.

The ordered pair \begin{align*}(9,3)\end{align*} is a solution for the first equation.

\begin{align*}(9,3)\end{align*}

Now, test the ordered pair in the second equation.

\begin{align*}x-y=4\end{align*}

Next, substitute \begin{align*}x=9\end{align*} and \begin{align*}y=3\end{align*} into the equation.

\begin{align*}\begin{array}{rcl} x-y &=& 4 \\ 9-3 &=& 4 \end{array}\end{align*}

Then, simplify the left side of the equation.

\begin{align*}\begin{array}{rcl} 9-3 &=& 4 \\ 6 &=& 4 \\ 6 & \neq & 4 \end{array}\end{align*}

Both sides of the equation are not equal.

The ordered pair \begin{align*}(9,3)\end{align*} is NOT a solution for the second equation.

Substitute the ordered pairs into the equation to see if it will make the statement true.

\begin{align*}(8,4)\end{align*}

First, begin with the first equation. 

\begin{align*}x+y=12\end{align*}

Next, substitute \begin{align*}x=8\end{align*} and \begin{align*}y=4\end{align*} into the equation.

\begin{align*}\begin{array}{rcl} x+y &=& 12 \\ 8+4 &=& 12 \end{array}\end{align*}

Then, simplify the left side of the equation.

\begin{align*}\begin{array}{rcl} 8+4 &=& 12 \\ 12 &=& 12 \end{array}\end{align*}

Both sides of the equation are equal.

The ordered pair \begin{align*}(8,4)\end{align*} is a solution for the first equation.

\begin{align*}(8,4)\end{align*}

Now, test the ordered pair in the second equation.

\begin{align*}x-y =4\end{align*}

Next, substitute \begin{align*}x=8\end{align*} and \begin{align*}y=4\end{align*} into the equation.

\begin{align*}\begin{array}{rcl} x-y &=& 4 \\ 8-4 &=& 4 \end{array}\end{align*}

Then, simplify the left side of the equation.

\begin{align*}\begin{array}{rcl} 8-4 &=& 4 \\ 4 &=& 4 \end{array}\end{align*}

Both sides of the equation are equal.

The ordered pair \begin{align*}(8,4)\end{align*} is a solution for the second equation.

The ordered pair \begin{align*}(8,4)\end{align*} is the solution for the system of linear equations since \begin{align*}x=8\end{align*} and \begin{align*}y=4\end{align*} makes both equations true.

\begin{align*}\begin{array}{rcl} x+y &=& 12 \\ x-y &=& 4 \end{array}\end{align*}

The solution can be written as \begin{align*}x=8\end{align*} and \begin{align*}y=4\end{align*} or as \begin{align*}\dbinom{x}{y}= \dbinom{8}{4}\end{align*}.

Let’s look at another system of linear equations.

\begin{align*}\begin{array}{rcl} x+y &=& -4 \\ 2x+2y &=& -8 \end{array}\end{align*}

Some ordered pairs that have a sum of \begin{align*}-4\end{align*} and that satisfy the first equation are \begin{align*}(-5,1) ,(-10,6)\end{align*} and \begin{align*}(7,-11)\end{align*} . Do these ordered pairs satisfy the second equation?

\begin{align*}(-5,1)\end{align*}

First, write down the second equation.

\begin{align*}2x+2y=-8\end{align*}

Next, substitute \begin{align*}x=-5\end{align*} and \begin{align*}y=1\end{align*} into the equation.

\begin{align*}\begin{array}{rcl} 2x+2y &=& -8 \\ 2(-5)+2(1) &=& -8 \end{array}\end{align*}

Then, simplify the left side of the equation.

\begin{align*}\begin{array}{rcl} 2(-5)+2(1) &=& -8 \\ -10+2 &=& -8 \\ -8 &=& -8 \end{array}\end{align*}

Both sides of the equation are equal.

The ordered pair \begin{align*}(-5,1)\end{align*} is a solution for the second equation.

\begin{align*}(-10,6)\end{align*}

First, write down the second equation.

\begin{align*}2x+2y=-8\end{align*}

Next, substitute \begin{align*}x=-10\end{align*}  and \begin{align*}y=6\end{align*} into the equation.

\begin{align*}\begin{array}{rcl} 2x+2y &=& -8 \\ 2(-10)+2(6) &=& -8 \end{array}\end{align*}

Then, simplify the left side of the equation.

\begin{align*}\begin{array}{rcl} 2(-10)+2(6) &=& -8 \\ -20+12 &=& -8 \\ -8 &=& -8 \end{array}\end{align*}

Both sides of the equation are equal.

The ordered pair \begin{align*}(-10,6)\end{align*} is a solution for the second equation.

\begin{align*}(7,-11)\end{align*}

First, write down the second equation.

\begin{align*}2x+2y=-8\end{align*}

Next, substitute \begin{align*}x=7\end{align*} and \begin{align*}y =-11\end{align*} into the equation.

\begin{align*}\begin{array}{rcl} 2x+2y &=& -8 \\ 2(7)+2(-11) &=& -8 \end{array}\end{align*}

Then, simplify the left side of the equation.

\begin{align*}\begin{array}{rcl} 2(7)+2(-11) &=& -8 \\ 14-22 &=& -8 \\ -8 &=& -8 \end{array}\end{align*}

Both sides of the equation are equal.

The ordered pair \begin{align*}(7,-11)\end{align*} is a solution for the second equation.

There are many ordered pairs that solve both equations in the system. In fact, there is an infinite number of solutions that solve both linear equations in this system. The second equation in the system is a multiple of the first equation.

\begin{align*}\begin{array}{rcl} x+y &=& -4 \\ 2(x+y &=& -4) \\ 2x+2y &=& -8 \end{array}\end{align*}

All systems of linear equations such that one equation is a multiple of the other will have an infinite number of solutions.

Let’s look at another system of linear equations.

\begin{align*}\begin{array}{rcl} x+y &=& 2 \\ x+y &=& 5 \end{array}\end{align*}

The ordered pair that would satisfy both of the equations in this system would have to add to give 2 and also add to give 5. There is no such ordered pair that would make both equations true.

Rewrite both equations in slope-intercept form.

First, subtract ‘\begin{align*}x\end{align*}’ from both sides of the equations to express the equations in the form \begin{align*}y=mx+b\end{align*}.

\begin{align*}\begin{array}{rcl} x+y &=& 2 \\ x-x+y &=& 2 -x \\ y &=& 2-x \end{array}\end{align*}

\begin{align*}\begin{array}{rcl} x+y &=& 5 \\ x-x+y &=& 5 -x \\ y &=& 5-x \end{array}\end{align*}

Then, rewrite the right side of the equation to match the slope-intercept form \begin{align*}y=mx+b\end{align*}.

\begin{align*}\begin{array}{rcl} y &=& mx+b \\ y &=& 2 -x \\ y &=& -x+2 \end{array}\end{align*}

\begin{align*}\begin{array}{rcl} y &=& mx+b \\ y &=& 5 -x \\ y &=& -x+5 \end{array}\end{align*}

Notice that both equations have the same slope of \begin{align*}m=-1\end{align*} but different \begin{align*}y\end{align*}-intercepts, \begin{align*}b=2\end{align*} and \begin{align*}b=5\end{align*}.

Lines having the same slope are parallel and parallel lines do not have any ordered pairs in common. Therefore, linear equations that have the same slope and different \begin{align*}y\end{align*}-intercepts will have no solution.

Examples

Example 1

Earlier, you were given a problem about Michelle and her piggy bank. She needs to figure out how many nickels and how many quarters she had in her bank. How can Michelle do this?

She can write a system of linear equations to model the information given and find the solution for the system of equations.

First, write down the information given in the problem.

The number of nickels plus the number of quarters equals 135 coins. The value of the nickels plus the value of the quarters equals $18.75.

Next, write the system of equations to model the given information.

Let \begin{align*}x= \text{the number of nickels }\end{align*}and let \begin{align*}y= \text{the number of quarters}\end{align*}.

\begin{align*}\begin{array}{rcl} x+y & = & 135 \\ 0.05x+0.25 y & = & 18.75 \end{array}\end{align*}

Next, write down ordered pairs that have a sum of 135.

\begin{align*}(65,70),(60,75),(70,65) \ \text{and} \ (75, 60)\end{align*}

All of these ordered pairs will satisfy the first equation. Which ordered pair will make the second equation true?

First, write down the second equation.

\begin{align*}0.05x+0.25 y = 18.75\end{align*}

Next, substitute \begin{align*}x=65\end{align*} and \begin{align*}y =70\end{align*} into the equation.

\begin{align*}\begin{array}{rcl} 0.05x+0.25y &=& 18.75 \\ 0.05(65)+0.25(70) &=& 18.75 \end{array}\end{align*}

Then, simplify the left side of the equation.

\begin{align*}\begin{array}{rcl} 0.05(65)+0.25(70) &=& 18.75 \\ 3.25 + 17.50 &=& 18.75 \\ 20.75 & \neq & 18.75 \end{array}\end{align*}

Both sides of the equation are NOT equal.

The ordered pair \begin{align*}(65,70)\end{align*} is NOT a solution for the second equation.

First, write down the second equation.

\begin{align*}0.05x+0.25 y = 18.75\end{align*}

Next, substitute \begin{align*}x=60\end{align*} and \begin{align*}y =75\end{align*} into the equation.

\begin{align*}\begin{array}{rcl} 0.05x+0.25y &=& 18.75 \\ 0.05(60)+0.25(75) &=& 18.75 \end{array}\end{align*}

Then, simplify the left side of the equation.

\begin{align*}\begin{array}{rcl} 0.05(60)+0.25(75) &=& 18.75 \\ 3.00 + 18.75 &=& 18.75 \\ 21.75 & \neq & 18.75 \end{array}\end{align*}

Both sides of the equation are NOT equal.

The ordered pair \begin{align*}(60,75)\end{align*} is NOT a solution for the second equation.

First, write down the second equation.

\begin{align*}0.05x+0.25 y = 18.75\end{align*}

Next, substitute \begin{align*}x=70\end{align*} and \begin{align*}y=65\end{align*} into the equation.

\begin{align*}\begin{array}{rcl} 0.05x+0.25y &=& 18.75 \\ 0.05(70)+0.25(65) &=& 18.75 \end{array}\end{align*}

Then, simplify the left side of the equation.

\begin{align*}\begin{array}{rcl} 0.05(70)+0.25(65) &=& 18.75 \\ 3.50 + 16.25 &=& 18.75 \\ 19.75 & \neq & 18.75 \end{array}\end{align*}

Both sides of the equation are NOT equal.

The ordered pair \begin{align*}(70,65)\end{align*} is NOT a solution for the second equation.

First, write down the second equation.

\begin{align*}0.05x+0.25 y = 18.75\end{align*}

Next, substitute \begin{align*}x=75\end{align*} and \begin{align*}y=60\end{align*} into the equation.

\begin{align*}\begin{array}{rcl} 0.05x+0.25y &=& 18.75 \\ 0.05(75)+0.25(60) &=& 18.75 \end{array}\end{align*}

Then, simplify the left side of the equation.

\begin{align*}\begin{array}{rcl} 0.05(75)+0.25(60) &=& 18.75 \\ 3.75 + 15.00 &=& 18.75 \\ 18.75 & = & 18.75 \end{array}\end{align*}

Both sides of the equation are equal.

The ordered pair \begin{align*}(75,60)\end{align*} is a solution for the second equation.

The solution is \begin{align*}\dbinom{x}{y}=\dbinom{75}{60}\end{align*}.

Michelle had 75 nickels and 60 quarters in her piggy bank.

Example 2

Which ordered pair makes both equations true?

\begin{align*}\begin{array}{rcl} x+y &=& 8 \\ 4x-y &=& -3 \end{array}\end{align*}

  1. \begin{align*}(2,6)\end{align*}
  2. \begin{align*}(3,15)\end{align*}
  3. \begin{align*}(4,4)\end{align*}
  4. \begin{align*}(1,7)\end{align*}

First, substitute \begin{align*}x =2\end{align*} and \begin{align*}y=6\end{align*} into both equations of the system.

\begin{align*}\begin{array}{rcl} x+y &=& 8 \\ 2+6 &=& 8 \end{array}\end{align*}

\begin{align*}\begin{array}{rcl} 4x-y &=& -3 \\ 4(2)-6 &=& -3 \end{array}\end{align*}

Next, simplify the left side of each of the equations.

\begin{align*}\begin{array}{rcl} 2+6 &=& 8 \\ 8 &=& 8 \end{array}\end{align*}

\begin{align*}\begin{array}{rcl} 4(2)-6 &=& -3 \\ 8-6 &=& -3 \\ 2 & \neq & -3 \end{array}\end{align*}

The ordered pair \begin{align*}(2,6)\end{align*} makes the first equation true but it does NOT make the second equation true.

First, substitute \begin{align*}x=3\end{align*} and \begin{align*}y=15\end{align*} into both equations of the system.

\begin{align*}\begin{array}{rcl} x+y &=& 8 \\ 3+15 &=& 18 \end{array}\end{align*}

\begin{align*}\begin{array}{rcl} 4x-y &=& -3 \\ 4(3)-15 &=& -3 \end{array}\end{align*}

Next, simplify the left side of each of the equations.

\begin{align*}\begin{array}{rcl} 3+15 &=& 8 \\ 18 & \neq & 8 \end{array}\end{align*}

\begin{align*}\begin{array}{rcl} 4(3)-15 &=& -3 \\ 12-15 &=& -3 \\ -3 &=& -3 \end{array}\end{align*}

The ordered pair \begin{align*}(3,15)\end{align*} does NOT make the first equation true but it does make the second equation true.

First, substitute \begin{align*}x=4\end{align*} and \begin{align*}y=4\end{align*} into both equations of the system.

\begin{align*}\begin{array}{rcl} x+y &=& 8 \\ 4+4 &=& 8 \end{array}\end{align*}

\begin{align*}\begin{array}{rcl} 4x-y &=& -3 \\ 4(4)-4 &=& -3 \end{array}\end{align*}

Next, simplify the left side of each of the equations.

\begin{align*}\begin{array}{rcl} 4+4 &=& 8 \\ 8 &=& 8 \end{array}\end{align*}

\begin{align*}\begin{array}{rcl} 4(4)-4 &=& -3 \\ 16 - 4 &=& -3 \\ 12 & \neq & -3 \end{array}\end{align*}

The ordered pair \begin{align*}(4,4)\end{align*} makes the first equation true but it does NOT make the second equation true.

First, substitute \begin{align*}x=1\end{align*} and \begin{align*}y=7\end{align*} into both equations of the system.

\begin{align*}\begin{array}{rcl} x+y &=& 8 \\ 1+7 &=& 8 \end{array}\end{align*}

\begin{align*}\begin{array}{rcl} 4x-y &=& -3 \\ 4(1)-7 &=& -3 \end{array}\end{align*}

Next, simplify the left side of each of the equations.

\begin{align*}\begin{array}{rcl} 1+7 &=& 8 \\ 8 &=& 8 \end{array}\end{align*}

\begin{align*}\begin{array}{rcl} 4(1)-7 &=& -3 \\ 4-7 &=& -3 \\ -3 &=& -3 \end{array}\end{align*}

The ordered pair \begin{align*}(1,7)\end{align*} makes the first equation true and it also makes the second equation true.

The solution is \begin{align*}\dbinom{x}{y}=\dbinom{1}{7}\end{align*}.

Example 3

Will the following system of linear equations have one solution, no solution or an infinite number of solutions? Justify your answer.

\begin{align*}\begin{array}{rcl} y &=& \frac{1}{2} x-3 \\ y &=& \frac{1}{2} x+2 \end{array}\end{align*}

First, write down what you know from the given equations.

The equations are written in the form \begin{align*}y=mx+b\end{align*}.

Both equations have the same slope of \begin{align*}\frac{1}{2}\end{align*}.

The equations do not have the same \begin{align*}y\end{align*}-intercepts.

Next, state the number of solutions for the system of equations.

Lines that are parallel have no ordered pairs in common so there will be no solution.

Example 4

Will the following system of linear equations have one solution, no solution or an infinite number of solutions? Justify your answer.

\begin{align*}\begin{array}{rcl} \frac{2}{3}x - \frac{3}{4}y &=& -1 \\ 8x-9y &=& -12 \end{array}\end{align*}

First, write down what you know from the given equations.

The second equation is a multiple of the first equation.

\begin{align*}\begin{array}{rcl} \overset{4} {\cancel {12}} \left( \frac{2}{\cancel{3}}x \right )- \overset{3} {\cancel {12}} \left( \frac{3}{\cancel{4}}y \right ) &=& 12(-1) \\ 4(2x)-3(3y) &=& 12(-1) \\ {\color{blue} 8x-9y }& {\color{blue}=}& {\color{blue}-12} \\ 8x-9y & = & -12 \end{array}\end{align*}

Next, state the number of solutions for the system of equations.

When one equation in a system of linear equations is a multiple of the other equation, there will be an infinite number of solutions for the system.

Example 5

Is the ordered pair \begin{align*}(3,-2)\end{align*} the solution for the following system of linear equations?

\begin{align*}\begin{array}{rcl} x-3y &=& 9 \\ 3x+y &=& 7 \end{array}\end{align*}

First, substitute \begin{align*}x=3\end{align*} and \begin{align*}y=-2\end{align*} into both equations of the system.

\begin{align*}\begin{array}{rcl} x-3y &=& 9 \\ 3-3(-2) &=& 9 \end{array}\end{align*}

\begin{align*}\begin{array}{rcl} 3x+y &=& 7 \\ 3(3)+(-2) &=& 7 \end{array}\end{align*}

Next, simplify the left side of each of the equations.

\begin{align*}\begin{array}{rcl} 3-3(-2) &=& 9 \\ 3+6 &=& 9 \\ 9 &=& 9 \end{array}\end{align*}

\begin{align*}\begin{array}{rcl} 3(3)+(-2) &=& 7 \\ 9-2 &=& 7 \\ 7 &=& 7 \end{array}\end{align*}

The ordered pair \begin{align*}(3,-2)\end{align*} makes the first equation true and it also makes the second equation true.

The solution is \begin{align*}\dbinom{x}{y}=\dbinom{3}{-2}\end{align*}.

Review

Figure out which pair is a solution for each given system.

1. Which ordered pair is a solution of the following system?

\begin{align*}\begin{array}{rcl} x-3y &=& 9 \\ 3x+y &=& 7 \end{array}\end{align*}

  1. \begin{align*}(6,-1)\end{align*}
  2. \begin{align*}(-1,-4)\end{align*}
  3. \begin{align*}(0,7)\end{align*}
  4. \begin{align*}(3,-2)\end{align*}

2. Which ordered pair is a solution of the following system?

\begin{align*}\begin{array}{rcl} && y =3x-7 \\ && 5x-3y =13 \end{array}\end{align*}

  1. \begin{align*}\left (3, \frac{2}{3} \right)\end{align*} 
  2. \begin{align*}(2,-1)\end{align*}
  3. \begin{align*}(4,7)\end{align*}
  4. \begin{align*}(5,8)\end{align*}

Determine whether each system has infinite solutions or no solutions.

3. \begin{align*}\begin{array}{rcl} x +y &=& 10 \\ y &=& -x +10 \end{array} \end{align*} 

4. \begin{align*}\begin{array}{rcl} 3x-6y &=& -24 \\ x-2y &=& -8 \end{array}\end{align*}

5. \begin{align*}\begin{array}{rcl} \frac{3}{4}x &=& \frac{2}{3}y-1 \\ 9x &=& 8y-12 \end{array} \end{align*}

6. \begin{align*}\begin{array}{rcl} y &=& 3x-5 \\ y &=& 3x-2 \end{array}\end{align*}

7. \begin{align*}\begin{array}{rcl} y &=& \frac{1}{2}x+3 \\ y &=& \frac{1}{2}x-2 \end{array} \end{align*}

Answer each question true or false.

8. Parallel lines have the same slope.

9. A linear system of equations cannot be graphed on the coordinate plane.

10. Parallel lines have infinite solutions.

11. Perpendicular lines have one solution.

12. Lines with an infinite number of solutions are not parallel.

13. Some linear systems do not have a solution.

14. To solve a linear system, you must have a value for \begin{align*}x\end{align*} and \begin{align*}y\end{align*}.

15. An ordered pair is never a solution for a linear system.

Review (Answers)

To see the Review answers, open this PDF file and look for section 9.13. 

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Vocabulary

system of equations

A system of equations is a set of two or more equations.

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