Solve the following equation 22r−1 2 2 r − 1 - 53 5 3 = 1r+2

Assessment: JAMB UTME - Mathematics - 1985 Subject: General Mathematics

Question 1 Report

Solve the following equation $\frac{2}{2 r - 1}$ - $\frac{5}{3}$ = $\frac{1}{r + 2}$

(

\frac{5}{2}

, 1)

(5, -4)

(2, 1)

(1,

\frac{- 5}{2}

)

(

\frac{- 5}{2}

, 1)

Answer Details

$\frac{2}{2 r - 1}$ - $\frac{5}{3}$ = $\frac{1}{r + 2}$
$\frac{2}{2 r - 1}$ - $\frac{1}{r + 2}$ = $\frac{5}{3}$
$\frac{2 r + 4 - 2 r + 1}{2 r - 1 (r + 2)}$ = $\frac{5}{3}$
$\frac{5}{(2 r + 1) (r + 2)}$ = $\frac{5}{3}$
5(2r - 1)(r + 2) = 15
(10r - 5)(r + 2) = 15
10r2 + 20r - 5r - 10 = 15
10r2 + 15r = 25
10r2 + 15r - 25 = 0
2r2 + 3r - 5 = 0
(2r2 + 5r)(2r + 5) = r(2r + 5) - 1(2r + 5)
(r - 1)(2r + 5) = 0
r = 1 or $\frac{- 5}{2}$

Read lesson note on Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

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Solve the following equation 22r−1 2 2 r − 1 - 53 5 3 = 1r+2

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