Solving Systems of Linear Equations in Two Variables

2.5.3 Solving Systems of Linear Equations in Two Variables

2.5.3 • Solving Systems of Linear Equations in Two Variables

Activity

Solve each system of equations without graphing, show your reasoning, and check your solutions.

System A

${\begin{matrix} 2 x + 3 y = 7 \\ - 2 x + 4 y = 14 \end{matrix}$

Use System A to answer questions 1 – 3.

What is the value of $x$ ?

-1

What is the value of $y$ ?

Show your reasoning and check the solution.

Compare your answer:

$\begin{matrix} \underset{―}{\begin{matrix} 2 x + 3 y & = & 7 \\ + - 2 x + 4 y & = & 14 \end{matrix}} \\ 7 y = 21 \\ y = 3 \end{matrix}$

$\begin{matrix} 2 x + 3 y & = & 7 \\ 2 x + 3 (3) & = & 7 \\ 2 x + 9 & = & 7 \\ 2 x & = & - 2 \\ x & = & - 1 \end{matrix}$

$\begin{matrix} 2 x + 3 y & = & 7 \\ 2 (- 1) + 3 (3) & = & 7 \\ - 2 + 9 & = & 7 \\ 7 & = & 7 ✓ \end{matrix}$

$\begin{matrix} - 2 x + 4 y & = & 14 \\ - 2 (- 1) + 4 (3) & = & 14 \\ 2 + 12 & = & 14 \\ 14 & = & 14 ✓ \end{matrix}$

System B

${\begin{matrix} 2 x + 3 y = 7 \\ 3 x - 3 y = 3 \end{matrix}$

Use System B to answer questions 4 – 6.

What is the value of $x$ ?

What is the value of $y$ ?

Show your reasoning and check the solution.

Compare your answer:

$\begin{matrix} \underset{―}{\begin{matrix} 2 x + 3 y & = & 7 \\ + 3 x - 3 y & = & 3 \end{matrix}} \\ 5 x = 10 \\ x = 2 \end{matrix}$

$\begin{matrix} 2 x + 3 y & = & 7 \\ 2 (2) + 3 y & = & 7 \\ 4 + 3 y & = & 7 \\ 3 y & = & 3 \\ y & = & 1 \end{matrix}$

$\begin{matrix} 3 x - 3 y & = & 3 \\ 3 (2) - 3 (1) & = & 3 \\ 6 - 3 & = & 3 \\ 3 & = & 3 ✓ \end{matrix}$

$\begin{matrix} 2 x + 3 y & = & 7 \\ 2 (2) + 3 (1) & = & 7 \\ 4 + 3 & = & 7 \\ 7 & = & 7 ✓ \end{matrix}$

System C

${\begin{matrix} 2 x + 3 y = 5 \\ 2 x + 4 y = 9 \end{matrix}$

Use System C to answer questions 7 – 9.

What is the value of $x$ ?

-3.5

What is the value of $y$ ?

Show your reasoning and check the solution.

Compare your answer:

$\begin{matrix} - 1 (2 x + 3 y & = & 5) \\ - 2 x - 3 y & = & - 5) \end{matrix}$

$\begin{matrix} \underset{―}{\begin{matrix} - 2 x - 3 y & = & - 5 \\ + 2 x + 4 y & = & 9 \end{matrix}} \\ y = 4 \end{matrix}$

$\begin{matrix} 2 x + 4 y & = & 9 \\ 2 x + 4 (4) & = & 9 \\ 2 x + 16 & = & 9 \\ 2 x & = & - 7 \\ x & = & - \frac{7}{2} \\ = & x - 3.5 \end{matrix}$

$\begin{matrix} 2 x + 3 y & = & 5 \\ 2 (- 3.5) + 3 (4) & = & 5 \\ - 7 + 12 & = & 5 \\ 5 & = & 5 ✓ \end{matrix}$

$\begin{matrix} 2 x + 4 y & = & 9 \\ 2 (- 3.5) + 4 (4) & = & 9 \\ - 7 + 16 & = & 9 \\ 9 & = & 9 ✓ \end{matrix}$

System D

${\begin{matrix} 6 x + 9 y = 48 \\ 6 x - 5 y = 20 \end{matrix}$

Use System D to answer questions 10 – 12.

10.

What is the value of $x$ ?

11.

What is the value of $y$ ?

12.

Show your reasoning and check the solution.

Compare your answer:

$\begin{matrix} - 1 (6 x + 9 y & = & 48) \\ - 6 x - 9 y & = & - 48 \end{matrix}$

$\begin{matrix} \underset{―}{\begin{matrix} - 6 x - 9 y & = & - 48 \\ + 6 x - 5 y & = & 20 \end{matrix}} \\ - 14 y = - 28 \\ y = 2 \end{matrix}$

$\begin{matrix} 6 x - 5 y & = & 20 \\ 6 x - 5 (2) & = & 20 \\ 6 x - 10 & = & 20 \\ 6 x & = & 30 \\ x & = & 5 \end{matrix}$

$\begin{matrix} 6 x + 9 y & = & 48 \\ 6 (5) + 9 (2) & = & 48 \\ 30 + 18 & = & 48 \\ 48 & = & 48 ✓ \end{matrix}$

$\begin{matrix} 6 x - 5 y & = & 20 \\ 6 (5) - 5 (2) & = & 20 \\ 30 - 10 & = & 20 \\ 20 & = & 20 ✓ \end{matrix}$

Are you ready for more?

Extending Your Thinking

This system has three equations:

${\begin{matrix} 3 x + 2 y - z = 7 \\ - 3 x + 5 y + 2 z = - 14 \\ 3 x + y - z = 10 \end{matrix}$

Add the first two equations to get a new equation.

Compare your answer:

$7 y + z = - 7$

Add the second two equations to get a new equation.

Compare your answer:

$6 y + z = - 4$

Solve the system of your two new equations, then answer questions 3 and 4.

What is the value of $y$ ?

Compare your answer:

-3

What is the value of $z$ ?

Compare your answer:

Solve the original system of equations, then answer questions 5 – 7.

What is the value of $x$ ?

Compare your answer:

What is the value of $y$ ?

Compare your answer:

-3

What is the value of $z$ ?

Compare your answer:

Self Check

A mechanic’s garage purchased 8 brake pads and 5 headlight sets for inventory and paid $1,460. Later, the mechanic purchased another order of 6 of the same brake pads and 5 of the same headlight sets and paid $1,220. How much does each part cost?

brake pads: $80; headlight sets: $148
brake pads: $80; headlight sets: $140
brake pads: $100; headlight sets: $100
brake pads: $120; headlight sets: $100

Additonal Resources

Solving a System of Equations by Elimination

Let’s reinforce the steps to solve a system of equations using elimination.

Step 1 - Write both equations in standard form. If any coefficients are fractions, clear them.

Step 2 - Check to see if the coefficients of one variable are opposites or equivalent.

Step 3 - Add or subtract the equations to eliminate one variable.

Step 4 - Solve for the remaining variable.

Step 5 - Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.

Step 6 - Write the solution as an ordered pair.

Step 7 - Check that the ordered pair is a solution to both original equations.

Example

Let’s look at an example. Solve the following system of equations using elimination:

${\begin{matrix} 3 x - 3 y = - 24 \\ - 3 x + 5 y = 30 \end{matrix}$

Step 1 - Write both equations in standard form. If any coefficients are fractions, clear them. ${\begin{matrix} 3 x - 3 y = - 24 \\ - 3 x + 5 y = 30 \end{matrix}$

Step 2 - Check to see if the coefficients of one variable are opposites or equivalent. The coefficients on the $x$ -variables are opposites. We can add the equations.

Step 3 - Add the equations to eliminate one variable. $\begin{matrix} 3 x - 3 y & = & - 24 \\ - 3 x + 5 y & = & 30 \\ 2 y & = & 6 \end{matrix}$

Step 4 - Solve for the remaining variable. $\begin{matrix} 2 y & = & 6 \\ y & = & 3 \end{matrix}$

Step 5 - Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable. \begin{array}{rcl}3x-3y&=&-24\\3x-3(3)&=&-24\\3x-9&=&-24\\3x&=&-15\\x&=&-5\end{array}

Step 6 - Write the solution as an ordered pair. $(- 5, 3)$

Step 7 - Check that the ordered pair is a solution to both original equations.

$\begin{matrix} 3 x - 3 y & = & - 24 \\ 3 (- 5) - 3 (3) & = & - 24 \\ - 15 - 9 & = & - 24 \\ - 24 & = & - 24 ✓ \end{matrix}$

$\begin{matrix} - 3 x + 5 y & = & 30 \\ - 3 (- 5) + 5 (3) & = & 30 \\ 15 + 15 & = & 30 \\ 30 & = & 30 ✓ \end{matrix}$

Try it

Solving a System of Equations by Elimination

Solve the following system of equations using elimination:

${\begin{matrix} 4 x - 6 y = 8 \\ x + 6 y = 17 \end{matrix}$

Compare your answer:

Here is how to solve this system of equations using elimination:

Step 1 - Write both equations in standard form. If any coefficients are fractions, clear them. $\begin{matrix} 4 x - 6 y & = & 8 \\ x + 6 y & = & 17 \end{matrix}$

Step 2 - Check to see if the coefficients of one variable are opposites or equivalent. The coefficients on the $y$ -variables are opposites. We can add the equations.

Step 3 - Add the equations to eliminate one variable. $\begin{matrix} 4 x - 6 y & = & 8 \\ x + 6 y & = & 17 \\ 5 x & = & 25 \end{matrix}$

Step 4 - Solve for the remaining variable. $\begin{matrix} 5 x & = & 25 \\ x & = & 5 \end{matrix}$

Step 5 - Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable. $\begin{matrix} 4 x - 6 y & = & 8 \\ 4 (5) - 6 y & = & 8 \\ 20 - 6 y & = & 8 \\ - 6 y & = & - 12 \\ y & = & 2 \end{matrix}$

Step 6 - Write the solution as an ordered pair. $(5, 2)$

Step 7 - Check that the ordered pair is a solution to both original equations.

$\begin{matrix} 4 x - 6 y & = & 8 \\ 4 (5) - 6 (2) & = & 8 \\ 20 - 12 & = & 8 \\ 8 & = & 8 ✓ \end{matrix}$

$\begin{matrix} x + 6 y & = & 17 \\ (5) + 6 (2) & = & 17 \\ 5 + 12 & = & 17 \\ 17 & = & 17 ✓ \end{matrix}$