1. Ch. 7 – Matrices and Systems of Equations
7.1 - Substitution

2. Plug it in for y in the other eqn., then solve for x!
Substitution
Ex: Solve the system by substitution.
The first eqn. is solved for y already!
Since y = x + 3, I can substitute x + 3 for y in the second equation!
After substituting, solve for x!
OK, so x=1. But we need an answer in the form of (x, y)…
I still need a y, so plug 1 in for x and solve for y!
You choose which equation to plug 1 back into.
y=4, so the answer is (1, 4)!
Verify by graphing!
This eqn. is solved for y!
Plug it in for y in the other eqn., then solve for x!

3. Solve the system by substitution:
(1, -1)
(-1, -2)
(2, -1/2)
(1/2, 4)
(-1, 1)

4. Solve the system: (1, 0) (-3, 16) (3, -8) (-1, 8) No solution
Infinite solutions

5. Ex: Solve the following system:
Uh-oh, we’re going to have a quadratic!
Solve the linear equation for y because the substitution will be easy!
Substitute to get x2 + 4x – (2x + 1) = 7
x2 + 2x – 8 = 0
To solve, factor or use quadratic formula!
(x + 4)(x – 2) = 0
x = -4 or x = 2
Plug them back in to get the y-values…
Answer: (-4, -7) and (2, 5)
Graph the equations to verify your answer!

6. Ex: Solve the following system:
Solve the linear equation for y because the substitution will be easy!
Substitute to get x2 + x + 4 = 3
x2 + x + 1 = 0
To solve, factor or use quadratic formula!
This equation doesn’t factor, and the quadratic formula gives a non-real answer, so…
No Solution!
Graph each equation to verify.

7. Solve the system by substitution:
(1, -1) and (-3, -9)
(0, -3) and (2, 1)
(3, 3)
(-1, -5) and (3, 3)
No solution

8. Solve the system by substitution:
(0, -5) and (4, 11)
(0, -5) and (1, -1)
(0, 5) and (3, 7)
(1, -1) and (4, 11)
No solution

9. Find the solution to the following system by graphing:
(2, 0)
(2.15, -0.23)
(0.18, -2.73), (1.67, -0.48)
(3.41, 0.88)
No solution