algebra 7.2 solving systems using substitution. you have already learned that the solution is the...
DESCRIPTION
There are two algebraic methods that allow you to solve a system easily without graphing. Today you will learn the method called SUBSTITUTION.TRANSCRIPT
AlgebraAlgebra7.2 7.2
Solving Systems Solving Systems Using SubstitutionUsing Substitution
You have already learned that the solution is the point of intersection of the two graphed lines.
Solution to a System of Linear Equations
To solve:1. Graph both equations2. Identify intersection point (x,y)3. Plug in to original equations to check
There are two algebraic methods that allow you to solve a system easily without graphing.
Today you will learn the method called SUBSTITUTION.
Steps1. In one equation, isolate one variable.2. Substitute expression from Step 1
into second equation and solve for the other variable.
3. Plug in value from Step 2 into revised equation from Step 1 and solve.
4. Check solution in both original equations.
Solve the linear system.3x + y = 52x – y = 10
Hint: It is usually easiest to isolate positive 1x or 1y.
y = 5 – 3x
2x – ( ) = 105 – 3x2x – 5 + 3x = 105x – 5 = 105x = 15x = 3
y = 5 – 3(3)y = 5 - 9y = - 4
The solution is (3, - 4)Check: 3(3) + (-4) = 5
2(3) – (-4) = 10
Solve the linear system.2x + 6y = 15x = 2y
2( ) + 6y = 152y
4y + 6y = 1510y = 15y = 15/10y = 3/2
x = 2( 3/2)x = 3
The solution is (3, 3/2)Check: 2(3) + 6(3/2) = 15
3 = 2(3/2)
x = 2y
You try!x + 2y = 4-x + y = -7 y = x – 7
x + 2( ) = 4x - 7
x + 2x - 14 = 43x – 14 = 43x = 18x = 6
y = 6 – 7y = -1
The solution is (6, - 1)Check: 6 + 2(-1) = 4
-6 + (-1) = -7
Mixture Problems
Systems are often used to solved mixture problems. These are problems when you mix two quantities. You know the total quantity and the total value, but not how much of each type. To solve:•Write one equation to describe
QUANTITY.•Write other equation to describe VALUE.
Set up a system and solve the mixture problem.An audio store sells two styles of I-pod Nanos. The 2 GB
model costs $150 and the 4GB model costs $225. Last Saturday the store sold 22 Nanos for a total of $3900. How many of each model did they sell?
Let x be the # of 2 GB models soldLet y be the # of 4 GB models sold
Quantity: x + y = 22Value: 150x + 225y = 3900
There were 14 2GB Nanos and 8 4 GB Nanos sold.
Now, you solve.
Homeworkpg. 408 #17 – 35 odd, 42