Objective: Solve systems of linear equations by substitution

Section 7-2 Solve Systems by SubstitutionSPI 23D: select the system of equations that could be used to solve a given real-world

problem

Objective:• Solve systems of linear equations by substitution

Three Methods of solving Systems of Equations:

• Solve by Graphing• Solve by Substitution• Solve by Elimination

Solve a System of Linear Equations by Substitution

Substitute: Replace a variable with an equivalent expression containing the other variable.

Solve the system of linear equations using substitution.

y = - 4x + 8y = x + 7

y = - 4x + 8

x + 7

1. Write an equation containing only one variable.

=

2. Solve the equation for x. x = 0.2

3. Substitute the x value into either equation to find y.

y = x + 7 y = 0.2 + 7 = 7.2

Substitute x + 7 for y.

6 + 8x = 28

Solve a System of Linear Equations by Substitution

Sometimes it is necessary to, first, solve one of the equations for a variable before using substitution.

Solve the system of linear equations using substitution.

6y + 8x = 283 = 2x - y

1. Solve one of the equations for a variable. 3 = 2x – y2x – 3 = y

2. Substitute the equation in step 1, into the remaining equation.

6y + 8x = 28(2x – 3)

3. Solve for x. Substitute x into either equation to find y.

x = 2.3 and y = 1.6

Real-world and Systems of Equations

Suppose you are thinking about buying a car. Car A cost $17,655 and you expect to pay an average of $1230 per year for fuel and repairs. Car B costs $15,900 and the average cost of fuel and repairs is $1425 per year. After how many years are the total costs for the cars the same?

1. Write two equations to model the problem.

C(y)= 1230y + 17,655 C(y)= 1425y + 15,900

2. Use substitution to solve.

1230y + 17,655 = 1425y + 15,90017,655 - 15,900 = 1425y - 1230y17,655 - 15,900 = 1425y - 1230y

1755 = 195y9 = y The cost will be the same after 9 years.