Question

A question on Khan Academy for PSAT prep. I don't get how that is equal to that.

Answer 1

Answer:

If you are solving for x or m, your answer will be

(-∞,∞)

Explanation:

Answer 2

$\frac{8}{3}-5mx=\frac{15mx-8}{-3}$

- This question is based on the fundamental concept of adding and subtracting fractions. I'll rewrite the question to see if this helps.

$\frac{8}{3}-\frac{5mx}{1}=\frac{15mx-8}{-3}$

- As you can see, every value in this equation is technically a fraction, but whenever a fraction has a denominator of $1$, it will often be shown without the denominator. With this idea in mind, let me introduce a new concept.

$\frac{8}{3}-\frac{5mx}{1}=\frac{15mx-8}{-3}\\\frac{8-5mx}{3-1}\neq\frac{15mx-8}{-3}$

- You cannot just subtract these two fractions' numerators and denominators together, you first need a common denominator. We do this by taking our fraction $\frac{5mx}{1}$, and multiplying it by the fraction $\frac{3}{3}$ ($\frac{3}{3}$ is the same as $1$ so it doesn't affect the equation to multiply by this fraction), then subtracting the two equations.

• We use the fraction $\frac{3}{3}$ because once multiplying $\frac{5mx}{1}$ by it, it will have a denominator the same as $\frac{8}{3}$.
• Remember that adding and subtracting fractions does not affect the denominator, that stays the same; it is the numerator that changes with addition and subtraction.

• ex.) $\frac{a}{c}+\frac{b}{c}=\frac{a+b}{c}$

• multiplying and dividing fractions affects both the numerator and denominator of each, and they can be multiplied or divided without worrying about having a common denominator.

• ex.) $\frac{a}{b}$ × $\frac{c}{d}=\frac{ac}{bd}$

$\frac{8}{3}-\frac{5mx}{1}=\frac{15mx-8}{-3}\\\frac{8}{3}-\frac{5mx}{1}(\frac{3}{3})=\frac{15mx-8}{-3}\\\frac{8}{3}-\frac{(3)5mx}{(3)1}=\frac{15mx-8}{-3}\\\frac{8}{3}-\frac{15mx}{3}=\frac{15mx-8}{-3}$

- Now that we have a common denominator, we can subtract these two fractions without worry.

$\frac{8}{3}-\frac{15mx}{3}=\frac{15mx-8}{-3}\\\frac{-15mx+8}{3}=\frac{15mx-8}{-3}\\(\frac{-1}{-1})\frac{-15mx+8}{3}=\frac{15mx-8}{-3}$

- I applied the same concept as multiplying $\frac{5mx}{1}$ by $\frac{3}{3}$ because $\frac{3}{3}$ won't affect the equation because it equals $1$. $\frac{-1}{-1}$ is the same as $1$, so by multiplying our fraction by this number, we aren't changing the fraction, just simplifying it really.

$(\frac{-1}{-1})\frac{-15mx+8}{3}=\frac{15mx-8}{-3}\\\frac{15mx-8}{-3}=\frac{15mx-8}{-3}$