Question

practice and problem solve each system by substitution. 8. \\(\

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

\\) 9. \\(\

$$\begin{cases}y = 2x + 10 \\\\ y = -2x - 6\\end{cases}$$

\\) 10. \\(\

$$\begin{cases}x + 2y = 8 \\\\ x + 3y = 12\\end{cases}$$

\\) 11. \\(\

$$\begin{cases}2x + 2y = 2 \\\\ -4x + 4y = 12\\end{cases}$$

\\) 12. \\(\

$$\begin{cases}y = 0.5x + 2 \\\\ -y = -2x + 4\\end{cases}$$

\\) 13. \\(\

$$\begin{cases}-x + y = 4 \\\\ 3x - 2y = -7\\end{cases}$$

\\) 4. \\(\

$$\begin{cases}3x + y = -8 \\\\ -2x - y = 6\\end{cases}$$

\\) 15. \\(\

$$\begin{cases}x + 2y = -1 \\\\ 4x - 4y = 20\\end{cases}$$

\\) 16. \\(\

$$\begin{cases}4x = y - 1 \\\\ 6x - 2y = -3\\end{cases}$$

\\)

Explanation:

Problem 8:

Step1: Substitute $y=x+3$ into second equation

$x+3=2x+4$

Step2: Solve for $x$

$x-2x=4-3 \implies -x=1 \implies x=-1$

Step3: Substitute $x=-1$ into $y=x+3$

$y=-1+3=2$

Problem 9:

Step1: Substitute $y=2x+10$ into second equation

$2x+10=-2x-6$

Step2: Solve for $x$

$2x+2x=-6-10 \implies 4x=-16 \implies x=-4$

Step3: Substitute $x=-4$ into $y=2x+10$

$y=2(-4)+10=-8+10=2$

Problem 10:

Step1: Isolate $x$ from first equation

$x=8-2y$

Step2: Substitute $x=8-2y$ into second equation

$8-2y+3y=12$

Step3: Solve for $y$

$8+y=12 \implies y=4$

Step4: Substitute $y=4$ into $x=8-2y$

$x=8-2(4)=8-8=0$

Problem 11:

Step1: Simplify first equation

$x+y=1 \implies x=1-y$

Step2: Substitute $x=1-y$ into second equation

$-4(1-y)+4y=12$

Step3: Expand and solve for $y$

$-4+4y+4y=12 \implies 8y=16 \implies y=2$

Step4: Substitute $y=2$ into $x=1-y$

$x=1-2=-1$

Problem 12:

Step1: Isolate $y$ from second equation

$y=2x-4$

Step2: Substitute $y=2x-4$ into first equation

$2x-4=0.5x+2$

Step3: Solve for $x$

$2x-0.5x=2+4 \implies 1.5x=6 \implies x=4$

Step4: Substitute $x=4$ into $y=2x-4$

$y=2(4)-4=8-4=4$

Problem 13:

Step1: Isolate $y$ from first equation

$y=x+4$

Step2: Substitute $y=x+4$ into second equation

$3x-2(x+4)=-7$

Step3: Expand and solve for $x$

$3x-2x-8=-7 \implies x=1$

Step4: Substitute $x=1$ into $y=x+4$

$y=1+4=5$

Problem 4:

Step1: Isolate $y$ from first equation

$y=-8-3x$

Step2: Substitute $y=-8-3x$ into second equation

$-2x-(-8-3x)=6$

Step3: Expand and solve for $x$

$-2x+8+3x=6 \implies x=6-8=-2$

Step4: Substitute $x=-2$ into $y=-8-3x$

$y=-8-3(-2)=-8+6=-2$

Problem 15:

Step1: Isolate $x$ from first equation

$x=-1-2y$

Step2: Substitute $x=-1-2y$ into second equation

$4(-1-2y)-4y=20$

Step3: Expand and solve for $y$

$-4-8y-4y=20 \implies -12y=24 \implies y=-2$

Step4: Substitute $y=-2$ into $x=-1-2y$

$x=-1-2(-2)=-1+4=3$

Problem 16:

Step1: Isolate $y$ from first equation

$y=4x+1$

Step2: Substitute $y=4x+1$ into second equation

$6x-2(4x+1)=-3$

Step3: Expand and solve for $x$

$6x-8x-2=-3 \implies -2x=-1 \implies x=\frac{1}{2}$

Step4: Substitute $x=\frac{1}{2}$ into $y=4x+1$

$y=4(\frac{1}{2})+1=2+1=3$

Answer:

1. $x=-1,\ y=2$
2. $x=-4,\ y=2$
3. $x=0,\ y=4$
4. $x=-1,\ y=2$
5. $x=4,\ y=4$
6. $x=1,\ y=5$
7. $x=-2,\ y=-2$
8. $x=3,\ y=-2$
9. $x=\frac{1}{2},\ y=3$