(C) 1041. If (3x – y) = (x + 5y) = 5:7, then thevalue of (x + y) : (x – y) is: About the author Mackenzie
In ratios, the numerator of LHS is equal to numerator of RHS and same with the denominators. Here, 3x – y = 5 —-(i) × 1 x + 5y = 7 —-(ii) × 3 Equation (i) becomes, 3x – y = 5 —-(a) Equation (ii) becomes, 3x + 15y = 21 —(b) Subtracting equation (a) from equation (b), we get, → 3x + 15y – (3x – y) = 21 – 5 → 3x + 15y – 3x + y = 16 → 16y = 16 → y = 1 From equation (ii), we have x = 7 – 5y x = 7 – 5(1) x = 7 – 5 = 2 Thus, [tex] \mapsto \: \tt \frac{x + y}{x – y} \\ \\ \mapsto \: \sf \frac{2 + 1}{2 – 1} = \frac{3}{1} \\ [/tex] Hence, (x + y) : (x – y) = 3 : 1 Reply
In ratios, the numerator of LHS is equal to numerator of RHS and same with the denominators.
Here,
Equation (i) becomes, 3x – y = 5 —-(a)
Equation (ii) becomes, 3x + 15y = 21 —(b)
Subtracting equation (a) from equation (b), we get,
→ 3x + 15y – (3x – y) = 21 – 5
→ 3x + 15y – 3x + y = 16
→ 16y = 16
→ y = 1
From equation (ii), we have
x = 7 – 5y
x = 7 – 5(1)
x = 7 – 5 = 2
Thus,
[tex] \mapsto \: \tt \frac{x + y}{x – y} \\ \\ \mapsto \: \sf \frac{2 + 1}{2 – 1} = \frac{3}{1} \\ [/tex]
Hence,