Question

unit 5: systems of equations and inequalities
name:
pd:

1. if $6a = 12b + 36$, what is $b$ in terms of $a$?

a. $\frac{1}{2}a - 3$
b. $\frac{1}{2}a + 3$
c. $a - \frac{3}{2}$
d. $\frac{a - 3}{2}$

1. sketch the two graphs below:

$\

$$\begin{cases} y = 6 - \\frac{1}{2}x \\\\ y = \\frac{5}{6}x - 2 \\end{cases}$$

$
(grid provided for sketching)

Explanation:

For Question 9:

Step1: Isolate the term with b

$6a - 36 = 12b$

Step2: Divide by 12 to solve for b

$b = \frac{6a - 36}{12}$

Step3: Simplify the fraction

$b = \frac{6(a - 6)}{12} = \frac{a}{2} - 3$

Step1: Identify key points for $y=6-\frac{1}{2}x$

• y-intercept (x=0): $y=6$, so point $(0,6)$
• x-intercept (y=0): $0=6-\frac{1}{2}x \implies x=12$, so point $(12,0)$

Step2: Identify key points for $y=\frac{5}{6}x-2$

• y-intercept (x=0): $y=-2$, so point $(0,-2)$
• x-intercept (y=0): $0=\frac{5}{6}x-2 \implies x=\frac{12}{5}=2.4$, so point $(2.4,0)$

Step3: Plot and connect points

Plot the two pairs of points on the grid, then draw a straight line through each pair.

Answer:

a. $\frac{1}{2}a - 3$

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For Question 10: