Question

If (3, 1) is the point of intersection of lines ax + by = 7 and bx + ay = 5, find the values of a and b.

Answer 1

(3, 1) is the point of intersection of lines ax + by = 7 and bx + ay = 5.
Thus, point (3, 1) satisfy both the equations.
i.e., ax + by = 7
$$\begin{gathered} \Rightarrow a\times 3+b\times 1=7 \\ \Rightarrow 3a+b=7...\left(1\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Also},bx+ay=5 \\ \Rightarrow b\times 3+a\times 1=5 \\ \Rightarrow 3b+a=5 \\ \Rightarrow a+3b=5...\left(2\right) \end{gathered}$$

Multiplying equation (2) by 3, we get:
3a + 9b = 15 ...(3)

Subtracting equation (1) from equation (3), we get:
3a + 9b = 15
3a + b = 7
$-$ $-$ $-$
--------------------------
8b = 8
⇒ b = 1
Substituting the value of b in equation (1), we get:

$$\begin{gathered} 3a+1=7 \\ \Rightarrow 3a=6 \\ \Rightarrow a=\frac{6}{3}=2 \\ \therefore a=2\mathrm{and}b=1 \end{gathered}$$