Question

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solving for x but you can’t combine like terms?
you can factor out a variable as a step in solving!
xy + 2x = 1 \t x(y + 2) = 1 \t x = \frac{1}{y+2}
yes, you can also divide an entire parenthesis.
solve for the given variable. show your work
e1) solve for x
12x + 7xp = 3b
e2) solve for a
5a + 3b = 2b + 3ab
e2) solve for n
7l + 10n = \frac{2lm - 5n}{m}

Explanation:

E1) Solve for $x$

Step1: Factor out $x$ from left side

$x(12 + 7p) = 3b$

Step2: Isolate $x$ by dividing both sides

$x = \frac{3b}{12 + 7p}$

E2) Solve for $a$

Step1: Rearrange terms with $a$ on left

$5a - 3ab = 2b - 3b$

Step2: Simplify right side, factor left $a$

$a(5 - 3b) = -b$

Step3: Isolate $a$ by dividing both sides

$a = \frac{-b}{5 - 3b}$ or $a = \frac{b}{3b - 5}$

E3) Solve for $N$

Step1: Multiply both sides by $M$ to eliminate denominator

$M(7L + 10N) = 2LM - 5N$

Step2: Expand the left side

$7LM + 10MN = 2LM - 5N$

Step3: Rearrange terms with $N$ on left

$10MN + 5N = 2LM - 7LM$

Step4: Simplify right side, factor left $N$

$N(10M + 5) = -5LM$

Step5: Isolate $N$ by dividing both sides

$N = \frac{-5LM}{10M + 5}$ or $N = \frac{-LM}{2M + 1}$ (simplified by dividing numerator/denominator by 5)

Answer:

E1) $x = \frac{3b}{12 + 7p}$
E2) $a = \frac{-b}{5 - 3b}$ (or $\frac{b}{3b - 5}$)
E3) $N = \frac{-LM}{2M + 1}$ (or $\frac{-5LM}{10M + 5}$)