Question

1. solve for $p$: $-5p + a - 10 = p - 5a$
2. find the least common multiple (lcm) of 21, 24, and 60.

graph these equations on a rectangular coordinate system:

1. $y = -2x + 4$
2. $y = 3$

Explanation:

Problem 12: Solve for $p$: $-5p + a - 10 = p - 5a$

Step1: Move $p$ terms to left side

Add $5p$ to both sides:
$a - 10 = 6p - 5a$

Step2: Move constant/ $a$ terms to left side

Add $5a$ and 10 to both sides:
$6a = 6p + 10$

Step3: Isolate $6p$ term

Subtract 10 from both sides:
$6a - 10 = 6p$

Step4: Solve for $p$

Divide by 6:
$p = \frac{6a - 10}{6} = a - \frac{5}{3}$

Step1: Prime factorize each number

$21 = 3 \times 7$
$24 = 2^3 \times 3$
$60 = 2^2 \times 3 \times 5$

Step2: Take highest powers of all primes

Highest powers: $2^3, 3^1, 5^1, 7^1$

Step3: Multiply the highest powers

$\text{LCM} = 2^3 \times 3 \times 5 \times 7$
$\text{LCM} = 8 \times 3 \times 5 \times 7 = 840$

1. Identify the equation is in slope-intercept form $y=mx+b$, where $m=-2$ (slope) and $b=4$ (y-intercept).
2. Plot the y-intercept at the point $(0, 4)$.
3. Use the slope: from $(0,4)$, move 1 unit right and 2 units down to get a second point $(1, 2)$, or 1 unit left and 2 units up to get $(-1, 6)$.
4. Draw a straight line through the plotted points.

Answer:

$p = a - \frac{5}{3}$

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Problem 13: Find the least common multiple (LCM) of 21, 24, and 60