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प्रश्न
Solve the following pair of equations:
43x + 67y = – 24, 67x + 43y = 24
योग
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उत्तर
Given pair of linear equations is
43x + 67y = – 24 ......(i)
And 67x + 43y = 24 ......(ii)
On multiplying equation (i) by 43 and equation (ii) by 67 and then subtracting both of them, we get
(67)2x + 43 × 67y = 24 × 67
(43)2x + 43 × 67y = – 24 × 43
– – +
{(67)2 – (43)2}x = 24(67 + 43)
⇒ (67 + 43)(67 – 43)x = 24 × 110 ......[∵ (a2 – b2) = (a – b)(a + b)]
⇒ 110 × 24x = 24 × 110
⇒ x = 1
Now, put the value of x in equation (i), we get
43 × 1 + 67y = – 24
⇒ 67y = – 24 – 43
⇒ 67y = – 67
⇒ y = – 1
Hence, the required values of x and y are 1 and –1, respectively.
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