Question

c. y = x² + 2

1. does x = 4 for the following equations? (show your work)

a. 4x + 2 = -2x - 10
b. 2x - 5 + 4x = -2x + 27
c. -32 = 8 - 10x

Explanation:

Step1: Substitute x = 4 into equation a

Substitute x = 4 into $4x + 2=-2x - 10$.
Left - hand side: $4\times4+2=16 + 2=18$.
Right - hand side: $-2\times4-10=-8 - 10=-18$.
Since $18
eq - 18$, x = 4 is not a solution for a.

Step2: Substitute x = 4 into equation b

Substitute x = 4 into $2x-5 + 4x=-2x + 27$.
Left - hand side: $2\times4-5+4\times4=8-5 + 16=19$.
Right - hand side: $-2\times4+27=-8 + 27=19$.
Since $19 = 19$, x = 4 is a solution for b.

Step3: Substitute x = 4 into equation c

Substitute x = 4 into $-32=8-10x$.
Right - hand side: $8-10\times4=8 - 40=-32$.
Since $-32=-32$, x = 4 is a solution for c.

Step4: Fill in the table for $y=x^{2}+2$

When $x=-4$, $y=(-4)^{2}+2=16 + 2=18$.
When $x=-3$, $y=(-3)^{2}+2=9 + 2=11$.
When $x=-2$, $y=(-2)^{2}+2=4 + 2=6$.
When $x=-1$, $y=(-1)^{2}+2=1 + 2=3$.
When $x = 0$, $y=0^{2}+2=2$.
When $x = 1$, $y=1^{2}+2=3$.
When $x = 2$, $y=2^{2}+2=4 + 2=6$.
When $x = 3$, $y=3^{2}+2=9 + 2=11$.
When $x = 4$, $y=4^{2}+2=16 + 2=18$.

Answer:

For the equations:
a. No
b. Yes
c. Yes
For the table:

IN (x)	OUT (y)
-3	11
-2	6
-1	3
0	2
1	3
2	6
3	11
4	18