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solve:x + 3y – 1 = 0; 2x + 2y + 1= 01. \(\rm x=\frac{5}{4};y=\frac{-3}{4}\)2. \(\rm x=\frac{-5}{4};y=\frac{3}{4}\)3. \(\rm x=\frac{-5}{4};y=\frac{-3}{4}\)4. \(\rm x=\frac{11}{4};y=\frac{3}{4}\) |
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Answer» Correct Answer - Option 2 : \(\rm x=\frac{-5}{4};y=\frac{3}{4}\) Given: x + 3y – 1 = 0; 2x + 2y + 1= 0 Calculation: x + 3y – 1 = 0 .....(1) 2x + 2y + 1= 0 ....(2) Multiplying by 2 in equation (1) ⇒ 2x + 6y – 2 = 0 ....(3) Now, subtracting the equation (2) from (3) ⇒ (2x + 6y – 2) – (2x + 2y + 1) = 0 ⇒ 4y – 3 = 0 ⇒ 4y = 3 ⇒ y = 3/4 Now, put the value of y in equation (1) we get, ⇒ x + 3 × (3/4) – 1 = 0 ⇒ x + 9/4 – 1 = 0 ⇒ x + 5/4 = 0 ⇒ x = -5/4 ∴ The value of x and y is -5/4 and 3/4 |
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