WAEC 2020 Mathematics · Question 22

Make m the subject of the relation k = $\sqrt{\frac{m - y}{m + 1}}$

Options

A.m = $\frac{y + k^2}{k^2 + 1}$
B.m = $\frac{y + k^2}{1 - k^2}$
C.m = $\frac{y - k^2}{k^2 + 1}$
D.m = $\frac{y - k^2}{1 - k^2}$

Answer: B. m = $\frac{y + k^2}{1 - k^2}$

Explanation written with AI and checked against the answer key

Step 1: Analyze the question We need to rearrange k = √((m − y)/(m + 1)) so that m stands alone on one side. Step 2: Concept Remove the square root by squaring both sides. Then clear the fraction, collect all the m terms on one side, and factorise. Step 3: Solution Square both sides: k² = (m − y)/(m + 1) Multiply both sides by (m + 1): k²(m + 1) = m − y Expand: k²m + k² = m − y Collect the m terms on one side: k² + y = m − k²m Factorise the right side: k² + y = m(1 − k²) Divide by (1 − k²): m = (y + k²)/(1 − k²) Step 4: Identify the answer This matches option B. Quick check: Let m = 3 and y = 1. Then k² = (3 − 1)/(3 + 1) = 1/2. Using our formula: (1 + 1/2)/(1 − 1/2) = (3/2)/(1/2) = 3. ✓ Step 5: Tip When m appears on both sides, move every m term to one side, factorise m out, then divide. Watch the signs: moving m to the other side is what gives (1 − k²) rather than (k² + 1). Options A and C are tempting traps for those who slip on this sign.

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Make m the subject of the relation k = m−ym+1\sqrt{\frac{m - y}{m + 1}}m+1m−y​​

Options

Answer: B. m = y+k21−k2\frac{y + k^2}{1 - k^2}1−k2y+k2​

Make m the subject of the relation k = $\sqrt{\frac{m - y}{m + 1}}$

Answer: B. m = $\frac{y + k^2}{1 - k^2}$