My Maths and I: WAEC Mathematics Question 2015 ( Essay Question Two 2a )

WAEC Mathematics Question 2015 ( Essay Question Two 2a )

May 18, 2020

By Kings

WAEC Mathematics 2015, WAEC Mathematics 2015 Essay

Solve the inequality:

Solution

$4+\frac{3}{4}(x+2)\leq \frac{3}{8}x+1$

Opening the bracket, we have;

$4+\frac{3x}{4}+\frac{6}{4}\leq \frac{3}{8}x+1$

Collecting like terms, we have;

$\frac{3x}{4}-\frac{3x}{8}\leq 1-4-\frac{6}{4}$

Taking the L.C.M. of both sides;

$\frac{6x-3x}{8}\leq \frac{4-16-6}{4}$

$\frac{3x}{8}\leq \frac{-18}{4}$

Multiplying both sides by 8 since x is required, we have;

$\frac{3x}{8}*8\leq \frac{-18}{4}*8$

$3x\leq -36$

Divide both side by 3 to find x

$\frac{3x}{3}\leq \frac{-36}{3}$

$\therefore x\leq -12$ ANSWERED

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1 comment:

Chibuike OzonoMay 19, 2020 at 8:53 AM
Nice one... I love this idea... Thank you for the clarity.
I have learnt.
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