Question

1. solve the equation $y = 3x - 5$ for $x$ when $y = 10$.
2. solve the equation $y = -2x + 7$ for $y$ when $x = 4$.
3. solve the equation $y = \frac{1}{2}x + 3$ for $x$ when $y = 8$.
4. write the equation in slope - intercept form for the line with a slope of $\frac{3}{2}$ and a y - intercept of - 4. then, determine if the point $(2, -1)$ satisfies the equation.
5. write the equation in slope - intercept form for a line that passes through the point $(3,5)$ and has a slope of 2. afterward, check if the point $(6,12)$ lies on the line.
6. write the equation in slope - intercept form for the line that passes through $(-2,1)$ and has a slope of - 3. then, verify if the point $(1, -8)$ is a solution to the equation.

Explanation:

Problem 8 Step1: Substitute y=10 into equation

$10 = 3x - 5$

Problem 8 Step2: Solve for x

$3x = 15 \Rightarrow x=5$

Problem9 Step1: Substitute x=4 into equation

$y = -2(4) + 7$

Problem9 Step2: Calculate y

$y = -8 + 7 = -1$

Problem10 Step1: Substitute y=8 into equation

$8 = \frac{1}{3}x + 3$

Problem10 Step2: Solve for x

$\frac{1}{3}x = 5 \Rightarrow x=15$

Problem11 Step1: Write slope-intercept form

$y = \frac{3}{2}x - 4$

Problem11 Step2: Check point (2,-1)

Left: $-1$, Right: $\frac{3}{2}(2)-4=3-4=-1$ (Equal, yes)

Problem12 Step1: Use point-slope form

$y - 5 = 2(x - 3)$

Problem12 Step2: Convert to slope-intercept

$y = 2x - 6 + 5 = 2x - 1$

Problem12 Step3: Check point (6,12)

Right: $2(6)-1=11
eq 12$ (No)

Problem13 Step1: Use point-slope form

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

Problem13 Step2: Convert to slope-intercept

$y = -3x -6 +1 = -3x -5$

Problem13 Step3: Check point (1,-8)

Right: $-3(1)-5=-8$ (Equal, yes)

Answer:

15
1
15
$y=\frac{3}{2}x - 4$; Yes
$y=2x - 1$; No
$y=-3x - 5$; Yes