The solution of the simultaneous equation \(\frac{X}{2}+\frac{y}{3}\)a

1.	The solution of the simultaneous equation \(\frac{X}{2}+\frac{y}{3}\)and x + y = 10 is given by (a) (6, 4) (b) (4, 6) (c) (–6, 4) (d) (6, –4)
Answer» (b) (4, 6) \(\frac{X}{2}\) + \(\frac{y}{3}\) = 4 \(\Rightarrow\) 6 x \(\frac{X}{2}\) + 6 x \(\frac{y}{3}\) = 6 x 4 \(\Rightarrow\) 3x + 2y = 24 ..........(i) Given, x + y = 10 \(\Rightarrow\) y = 10 – x .....(ii) Substituting the value of y in (i), we get 3x + 2 (10 – x) = 24 \(\Rightarrow\)3x + 20 – 2x = 24 \(\Rightarrow\) x = 24 – 20 = 4 \(\therefore\) From eqn (ii), y = 10 – 4 = 6.

The solution of the simultaneous equation \(\frac{X}{2}+\frac{y}{3}\)and x + y = 10 is given by (a) (6, 4) (b) (4, 6) (c) (–6, 4) (d) (6, –4)

Answer»

(b) (4, 6)

\(\frac{X}{2}\) + \(\frac{y}{3}\) = 4

\(\Rightarrow\) 6 x \(\frac{X}{2}\) + 6 x \(\frac{y}{3}\) = 6 x 4

\(\Rightarrow\) 3x + 2y = 24 ..........(i)

Given, x + y = 10 \(\Rightarrow\) y = 10 – x .....(ii)

Substituting the value of y in (i), we get

3x + 2 (10 – x) = 24

\(\Rightarrow\)3x + 20 – 2x = 24

\(\Rightarrow\) x = 24 – 20 = 4

\(\therefore\) From eqn (ii), y = 10 – 4 = 6.