Question

1. Solve. 3(h – 4) = –1/2(24 – 6h) 2. Solve for x. ax + bx = –c

Answer 1

ANSWER TO QUESTION 1

$3(h-4) = - \frac{1}{2} (24 - 6h)$

We multiply through by the Least Common Multiple which is 2.

$2 \times 3(h-4) = - 2 \times \frac{1}{2} (24 - 6h)$

$6(h - 4) = - 1(24 - 6h)$

We expand brackets to obtain,

$6h - 24 = - 24 + 6h$

Grouping like terms, have

$6h - 6h = - 24 + 24$

$\Rightarrow 0 = 0$

Whenever you solve an equation and you get the above result, you don't have to get confuse.

It simply means the question does not have a UNIQUE solution.

That any real number will number will satisfy the above equation.

Hence,

$h \in R$




ANSWER TO QUESTION 2

$ax + bx = - c$

We factor x to obtain;

$x(a + b) = - c$

We divide both sides by (a+b)

$x = - \frac{ c}{a + b}$