How do I solve algebraic fractions with two fractions on one side?
$$\frac{a}{4}-\frac{a+2}{3}=9$$
Just like that one, what would I do? Please explain why you did what you did. I can't show working out for something I don't know how to do.
$\endgroup$ 53 Answers
$\begingroup$Step 1: Find the common denominator: $3\cdot 4 = 12$
Step 2: Multiply both sides by common denominator $12$.
$12\cdot \left(\dfrac{a}{4} - \dfrac{a+2}{3}\right) = 12\cdot 9$.
Step 3: Distribute $12$ into the parentheses of the left side.
$12\cdot \dfrac{a}{4} - 12\cdot \dfrac{a+2}{3} = 108$, and simplify.
$3a - 4(a+2) = 108$, and simplify again.
$3a - 4a - 8 = 108$, and simplify again.
$-a - 8 = 108$, add $8$ both sides.
$-a - 8 + 8 = 108 + 8$, simplify.
$-a = 116$, and finally multiply $-1$ to both sides to get the answer.
$a = -116$.
$\endgroup$ $\begingroup$Multiply the entire equation by all of the denominators, and then solve for A as if there were no fractions.
So in this case, multiply the entire equation by 4. Then multiply by 3.
$\endgroup$ $\begingroup$$$\frac {a}{4} = \frac{1} {4}a$$ $$\frac {a+2} {3} = \frac {1}{3}a + \frac {2} {3}$$
$$\frac {1} {4}a-(\frac {1} {3}a + \frac {2} {3}) = 9$$ $$\frac{1}{4}a - \frac {1} {3}a - \frac {2}{3} = 9$$
Let's find the value of $\frac {1}{4}a -\frac {1}{3}a$. Then we will subtract the results by $\frac {2} {3}$ or add $-\frac {2}{3}$ $$\frac {1} {4}a -\frac{1}{3}a = \frac {3 - 4}{12}a = -\frac {1} {12}a$$ So: $$-\frac {1} {12}a - \frac {2} {3} = 9$$ $$-\frac {1} {12}a = 9\frac {2} {3}$$ $$-\frac {1} {12}a = \frac {29} {3}$$ $$a = \frac {29} {3} * -\frac {12} {1}$$ $$a = \frac {29 * -12}{3 * 1}$$ $$a = \frac {-348} {3}$$ $$a = -116$$
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