Equations with brackets - example4

Where brackets are involved in equations, the brackets are first removed before solving the equations further.

To remove brackets from an equation, multiply the terms in the bracket by the number attached to the bracket.

For instance, if we have 2(x+1) and we are to remove bracket, we multiply x+1 by 2 as follows

2*x + 2*1 to become 2x + 2

Question

Solve the following equation 3(2q – 7) = 2(q + 1) – 7

Solution

Given the equation 3(2q – 7) = 2(q + 1) – 7

The first step here is to remove brackets (ie open the brackets)

3(2q – 7) = 2(q + 1) – 7

3*2q – 3*7 = 2*q + 2*1 – 7 and this becomes

3*2q – 3*7 = 2*q + 2*1 – 7

6q – 21 = 2q + 2 – 7

We can now talk of like terms.

The like terms in the equation 6q and 2q as well as -21, 2 and -7

Thus, we say bring like terms together as follows

6q – 2q = 2 - 7 + 21

You noticed in the above that +2q in the right hand side (RHS) crosses to the left hand side (LHS) to become -2q

Also, -21 in the left hand side (LHS) crosses the equality sign to the right hand side (RHS) to become +21

6q – 2q = 2 - 7 + 21

4q = 16

The coefficient of q is 4. To get q, we now divide both sides by 4 (coefficient of q) as follows

^4q/₄ = ¹⁶/₄

q = 4

Check for correctness

The given equation is 3(2q – 7) = 2(q + 1) – 7

LHS =   3(2q – 7)

     =   3(2*4 – 7)

=   3(8 – 7)

=   3 * 1

=   3

RHS =   2(q + 1) – 7

     =   2(4 + 1) – 7

=   2(5) – 7

=   2*5 – 7

=   10 – 7

     =   3 (Same as LHS)