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Algebra 1

2.5.3 Solving Systems of Linear Equations in Two Variables

Algebra 12.5.3 Solving Systems of Linear Equations in Two Variables

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Activity

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

System A

{2x+3y=72x+4y=14{2x+3y=72x+4y=14

Use System A to answer questions 1 – 3.

1.

What is the value of xx?

2.

What is the value of yy?

3.

Show your reasoning and check the solution.

System B

{2x+3y=73x3y=3{2x+3y=73x3y=3

Use System B to answer questions 4 – 6.

4.

What is the value of xx?

5.

What is the value of yy?

6.

Show your reasoning and check the solution.

System C

{2x+3y=52x+4y=9{2x+3y=52x+4y=9

Use System C to answer questions 7 – 9.

7.

What is the value of xx?

8.

What is the value of yy?

9.

Show your reasoning and check the solution.

System D

{6x+9y=486x5y=20{6x+9y=486x5y=20

Use System D to answer questions 10 – 12.

10.

What is the value of xx?

11.

What is the value of yy?

12.

Show your reasoning and check the solution.

Are you ready for more?

Extending Your Thinking

This system has three equations:

{3x+2yz=73x+5y+2z=143x+yz=10{3x+2yz=73x+5y+2z=143x+yz=10

1.

Add the first two equations to get a new equation.

2.

Add the second two equations to get a new equation.

3.

What is the value of yy?

4.

What is the value of zz?

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

5.

What is the value of xx?

6.

What is the value of yy?

7.

What is the value of zz?

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?
  1. brake pads: $80; headlight sets: $148
  2. brake pads: $80; headlight sets: $140
  3. brake pads: $100; headlight sets: $100
  4. 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:

{3x3y=243x+5y=30{3x3y=243x+5y=30

Step 1 - Write both equations in standard form. If any coefficients are fractions, clear them. {3x3y=243x+5y=30{3x3y=243x+5y=30

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

Step 3 - Add the equations to eliminate one variable. 3x3y=243x+5y=302y=63x3y=243x+5y=302y=6

Step 4 - Solve for the remaining variable. 2y=6y=32y=6y=3

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)(5,3)

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

3x3y=243(5)3(3)=24159=2424=243x3y=243(5)3(3)=24159=2424=24

3x+5y=303(5)+5(3)=3015+15=3030=303x+5y=303(5)+5(3)=3015+15=3030=30

Try it

Try It: Solving a System of Equations by Elimination

Solve the following system of equations using elimination:

{4x6y=8x+6y=17{4x6y=8x+6y=17

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