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) = - (1)/(2) (24 - 6h)`

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


`2 * 3(h-4) = - 2 * (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 = - ( c)/(a + b)`