Question

algebra 2 - unit 1 part 2 - section 1.6 mini quiz
name
solve each by graphing. state the solution (or solutions) or a point within the solution.

1. $y=\frac{1}{2}x + 2$

$y=-|x| + 2$
solution:

1. $y=\frac{2}{1}x + 3$

$y + 2x=-5$
solution:

1. $-x + 2=|x| - 4$

solution:

1. $\frac{1}{2}x + 5=-x + 2$

solution:

1. $y < 3$

$y > x + 5$
solution:

1. $y > 2x + 2$

$y < |x - 5|$
solution:

1. the system was graphed below without the shading. shade correctly to find the solution section and then state one solution.

$y < -\frac{2}{3}x + 7$
$y > x + 2$
solution:

1. the inequality is being solve by graphing. hightlight the section of the graph that is the solution set of the inequality and then state the solution.

$\frac{3}{2}|x - 6| + 1 < \frac{1}{2}x + 8$
solution:

Explanation:

Step1: Verify Q1 intersection

Set $\frac{2}{3}x + 2 = -|x| + 2$
$\frac{2}{3}x = -|x|$
Test $x=0$: $0=0$, $y=2$. The marked $(2,5)$ is incorrect; correct solution is $(0,2)$.

Step2: Verify Q2 intersection

Rewrite $y+2x=-5$ as $y=-2x-5$. Set $\frac{2}{3}x + 3 = -2x -5$
$\frac{2}{3}x + 2x = -5 -3$
$\frac{8}{3}x = -8$
$x=-3$, $y=-2(-3)-5=1$. Solution $(-3,1)$ is correct.

Step3: Verify Q3 intersection

Set $-x+2 = |x| -4$
Case1: $x\geq0$: $-x+2=x-4$ → $2x=6$ → $x=3$, $y=-3+2=-1$.
Case2: $x<0$: $-x+2=-x-4$ → $2=-4$ (no solution). Solution $(3,-1)$ is correct.

Step4: Verify Q4 intersection

Set $\frac{1}{2}x + 5 = -x + 2$
$\frac{1}{2}x + x = 2 -5$
$\frac{3}{2}x = -3$
$x=-2$, $y=-(-2)+2=4$. The marked $(-11,1)$ is incorrect; correct solution is $(-2,4)$.

Step5: Verify Q5 solution region

$y<3$ (shade below $y=3$) and $y>x+5$ (shade above $y=x+5$). The region $x+5

Step6: Verify Q6 solution region

$y>2x+2$ (shade above line) and $y<|x-5|$ (shade below V-shape). The marked $(0,2)$ is on $y=2x+2$, not in the region; a valid point is $(0,1)$.

Step7: Shade Q7 region

$y<-\frac{2}{3}x+7$ (shade below dashed line), $y>x+2$ (shade above dashed line). The overlapping region is between the lines. A valid point is $(0,3)$.

Step8: Shade Q8 region

$\frac{3}{2}|x-6| + 1 < \frac{1}{2}x + 8$ → $3|x-6| < x + 14$
Case1: $x\geq6$: $3x-18 < x+14$ → $2x<32$ → $x<16$
Case2: $x<6$: $-3x+18 < x+14$ → $4x>4$ → $x>1$
Solution is $1

Answer:

1. Correct solution: $\boldsymbol{(0,2)}$
2. Correct solution: $\boldsymbol{(-3,1)}$
3. Correct solution: $\boldsymbol{(3,-1)}$
4. Correct solution: $\boldsymbol{(-2,4)}$
5. Valid point in region: $\boldsymbol{(-6,0)}$
6. Valid point in region: $\boldsymbol{(0,1)}$
7. Valid point in region: $\boldsymbol{(0,3)}$
8. Valid point in region: $\boldsymbol{(6,7)}$