Question

solve each system by graphing. check your answers.

1. \\(\

$$\begin{cases}y = x - 2\\\\x + y = 10\\end{cases}$$

\\)

1. \\(\

$$\begin{cases}y = 7 - x\\\\-5x + y = 1\\end{cases}$$

\\)

1. \\(\

$$\begin{cases}y = x^2 + 5x - 2\\\\y = 3x - 2\\end{cases}$$

\\)

1. \\(\

$$\begin{cases}y = -x^2 + 2x + 18\\\\y = 5x - 10\\end{cases}$$

\\)

1. \\(\

$$\begin{cases}y = -x^2 + 2x - 3\\\\y = x^2 + 4x - 3\\end{cases}$$

\\)

1. \\(\

$$\begin{cases}y = -x\\\\y = |x + 2|\\end{cases}$$

\\)
solve by graphing on the calculator only. do not graph by hand.

1. \\(\

$$\begin{cases}y = x^2 + 2x - 3\\\\y = -x^2 - 2x + 3\\end{cases}$$

\\)

1. \\(\

$$\begin{cases}y = 2x + 4\\\\y = -5x - 3\\end{cases}$$

\\)

1. \\(\

$$\begin{cases}x - y = 2\\\\y = -x\\end{cases}$$

\\)

1. \\(\

$$\begin{cases}y = 2x^2 - 4\\\\y = x^2 - 4x + 1\\end{cases}$$

\\)

1. \\(\

$$\begin{cases}y = -x^2 + x + 2\\\\y = x^2 - 3x - 4\\end{cases}$$

\\)

1. \\(\

$$\begin{cases}y = -2x - 2\\\\2y = x + 6\\end{cases}$$

\\)

Explanation:

Step1: Solve system 1 via substitution

Substitute $y=x-2$ into $x+y=10$:
$x+(x-2)=10$
$2x-2=10$
$2x=12$
$x=6$
Substitute $x=6$ into $y=x-2$:
$y=6-2=4$

Step2: Solve system 2 via substitution

Substitute $y=7-x$ into $-5x+y=1$:
$-5x+(7-x)=1$
$-6x+7=1$
$-6x=-6$
$x=1$
Substitute $x=1$ into $y=7-x$:
$y=7-1=6$

Step3: Solve system 3 via substitution

Set $x^2+5x-2=3x-2$:
$x^2+2x=0$
$x(x+2)=0$
$x=0$ or $x=-2$
For $x=0$: $y=3(0)-2=-2$
For $x=-2$: $y=3(-2)-2=-8$

Step4: Solve system 4 via substitution

Set $-x^2+2x+18=5x-10$:
$-x^2-3x+28=0$
$x^2+3x-28=0$
$(x+7)(x-4)=0$
$x=-7$ or $x=4$
For $x=-7$: $y=5(-7)-10=-45$
For $x=4$: $y=5(4)-10=10$

Step5: Solve system 5 via substitution

Set $-x^2+2x-3=x^2+4x-3$:
$-2x^2-2x=0$
$-2x(x+1)=0$
$x=0$ or $x=-1$
For $x=0$: $y=0^2+4(0)-3=-3$
For $x=-1$: $y=(-1)^2+4(-1)-3=-6$

Step6: Solve system 6 via substitution

Set $-x=|x+2|$
Case 1: $x+2\geq0$ ($x\geq-2$):
$-x=x+2$
$-2x=2$
$x=-1$ (valid, since $-1\geq-2$)
$y=-(-1)=1$
Case 2: $x+2<0$ ($x<-2$):
$-x=-(x+2)$
$-x=-x-2$
$0=-2$ (no solution)

Step7: Solve system 7 via substitution

Set $x^2+2x-3=-x^2-2x+3$:
$2x^2+4x-6=0$
$x^2+2x-3=0$
$(x+3)(x-1)=0$
$x=-3$ or $x=1$
For $x=-3$: $y=(-3)^2+2(-3)-3=0$
For $x=1$: $y=1^2+2(1)-3=0$

Step8: Solve system 8 via substitution

Set $2x+4=-5x-3$:
$7x=-7$
$x=-1$
Substitute $x=-1$ into $y=2x+4$:
$y=2(-1)+4=2$

Step9: Solve system 9 via substitution

Substitute $y=-x$ into $x-y=2$:
$x-(-x)=2$
$2x=2$
$x=1$
$y=-1$

Step10: Solve system 10 via substitution

Set $2x^2-4=x^2-4x+1$:
$x^2+4x-5=0$
$(x+5)(x-1)=0$
$x=-5$ or $x=1$
For $x=-5$: $y=2(-5)^2-4=46$
For $x=1$: $y=2(1)^2-4=-2$

Step11: Solve system 11 via substitution

Set $-x^2+x+2=x^2-3x-4$:
$-2x^2+4x+6=0$
$x^2-2x-3=0$
$(x-3)(x+1)=0$
$x=3$ or $x=-1$
For $x=3$: $y=-(3)^2+3+2=-4$
For $x=-1$: $y=-(-1)^2+(-1)+2=0$

Step12: Solve system 12 via substitution

Substitute $y=-2x-2$ into $2y=x+6$:
$2(-2x-2)=x+6$
$-4x-4=x+6$
$-5x=10$
$x=-2$
Substitute $x=-2$ into $y=-2x-2$:
$y=-2(-2)-2=2$

Answer:

1. $(6, 4)$
2. $(1, 6)$
3. $(0, -2)$ and $(-2, -8)$
4. $(-7, -45)$ and $(4, 10)$
5. $(0, -3)$ and $(-1, -6)$
6. $(-1, 1)$
7. $(-3, 0)$ and $(1, 0)$
8. $(-1, 2)$
9. $(1, -1)$
10. $(-5, 46)$ and $(1, -2)$
11. $(3, -4)$ and $(-1, 0)$
12. $(-2, 2)$