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Integrated Algebra
NOTES:
Solving systems – substitution
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
If it is not possible to graph two equations to find the solution, you
can find the solution algebraically.

When you use substitution to solve a system, you are creating an
equation in one variable.
To solve a system this way, follow these steps:
1) Pick one equation and isolate either x or y. Look for a
variable with a coefficient of 1.
2) Substitute for that variable in the other equation. Solve.
3) Take the value you just found and substitute it into either
equation to find the value of the other variable.
4) Write the solution as a coordinate pair.
5)
EX: Solve the
6)following system of equations:
y = 3x + 2
2x + y = -8
1) Isolate one variable. Since the first equation is already isolated for
y, use this to substitute for y in the other equation.
2) Substitute:
Since y = 3x + 2
2x +
y
= -8
2x + 3x + 2 = -8
5x + 2 = -8
5x = -10
x = -2
3) Substitute -2 in for x in either equation to find y.
y = 3 (-2) + 2
= -6 + 2
= -4
so,
x = -2 and y = -4
4) Solution: This means that these two lines will intersect at (-2, -4)
Integrated Algebra
EX: Solve the following system of equations by substitution.
y = 3x – 10
y = 2x - 5
1) Isolate one of the variables
2) Substitute into other equation
3) Substitute to solve for other variable.
4) Solution
TRY: Solve the system of equations by substitution.
3x + 5y = 2
x + 4y = -4
1) Isolate one of the variables.
2) Substitute it into the other equation.
3) Substitute to solve for the other one.
4) Solution.
Integrated Algebra
HOMEWORK: Solving systems by substitution
Solve each system by substitution>
1. y = x – 7
2x + y = 8
2. y = x - 3
y = -3x + 25
3. x + 2y = 200
x = y + 50
4. 2x + 4y = -6
x – 3y = 7
5. 5x – 3y = -4
x + y = -4
6. x + y = 0
x=y+4
7. 2x + 5y = -6
4x + y = -12
8. y = 5x + 5
y = 15x – 1
9. -5x + y = 6
2x – 3y = 60
10. 3x – 4y = -5
x=y+2
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