Transcript 7-2 PPT Solve Systems Substitution

Section 7-2 Solve Systems by Substitution
SPI 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 + 8
y= x+7
1. Write an equation containing only one variable.
y = - 4x + 8
Substitute x + 7 for y.
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
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 = 28
3 = 2x - y
1. Solve one of the equations for a variable. 3 = 2x – y
2x – 3 = y
2. Substitute the equation in step 1, into the remaining equation.
6y + 8x = 28
6(2x – 3) + 8x = 28
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 + 15,900
17,655 - 15,900 = 1425y - 1230y
17,655 - 15,900 = 1425y - 1230y
1755 = 195y
9=y
The cost will be the same after 9 years.