Solve the following pair of linear equations by the substitution and cross-multiplication methods: $8 x + 5 y = 9, 3 x + 2 y = 4$

x = - 3, y = 1

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x = 7, y = - 6

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x = - 2, y = 5

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x = 1, y = - 6

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Solution

The correct option is C $x = - 2, y = 5$
Given system of Equations are:
$8 x + 5 y = 9$ ....(1)
and $3 x + 2 y = 4$ ....(2)
Now rearranging both the equation
$8 x + 5 y - 9 = 0.. (3)$
$3 x + 2 y - 4 = 0... (4)$
Now from (3) and (4)
$a_{1} = 8, b_{1} = 5, c_{1} = - 9$
$a_{2} = 3, b_{2} = 2, c_{2} = - 4$
Now putting formula for cross multiplication
$\frac{x}{b_{1} c_{2} - b_{2} c_{1}} = \frac{y}{c_{1} a_{2} - c_{2} a_{1}} = \frac{1}{a_{1} b_{2} - a_{2} b_{1}}$
$\Rightarrow \frac{x}{5 (- 4) - 2 (- 9)} = \frac{y}{- 9 (3) - (- 4) 8} = \frac{1}{8 (2) - 3 (5)}$
$\Rightarrow \frac{x}{- 20 + 18} = \frac{y}{- 27 + 32} = \frac{1}{16 - 15}$
$\Rightarrow \frac{x}{- 2} = \frac{y}{5} = \frac{1}{1}$
$\Rightarrow x = - 2$
$\Rightarrow y = 5$

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Similar questions

Solve the following pair of linear equations by the elimination method and the substitution method :

(i) x+y=5 and 2x−3y=4
(ii) 3x+4y=10 and 2x−2y=2
(iii) 3x−5y−4=0 and 9x=2y+7
(iv) x2+2y3=−1 and x−y3=3

\frac{3 x}{2} - \frac{5 y}{3} = - 2; \frac{x}{3} + \frac{y}{2} = \frac{13}{6}

Question 1 (i)
Solve the following pair of linear equations by the elimination method and the substitution method:
x + y =5 and 2x - 3y = 4

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$7 y - x = 8$

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