Question

Solve the following system of equations using the cross multiplication method.

$$\begin{gathered} ax+by=a-b \\ bx-ay=a+b \end{gathered}$$

Answer 1

Step 1:Analysing the given equations

The given system of equations may be written as

$$\begin{gathered} ax+by-(a-b)=0 \\ bx-ay-(a+b)=0 \end{gathered}$$

The given system of equation is of the forms

$$\begin{gathered} a_{1}x+b_{1}y+c_1=0 \\ {a}_{2}x+b_{2}y+c_2=0 \end{gathered}$$

Step 2: using cross multiplication method

where, $$\begin{gathered} a_1=a,b_1=b,c_1=-(a-b) \\ {a}_2=b,b_2=-a,c_2=-(a+b) \end{gathered}$$

Substitute the given values in the general form of equation, we get

$$\begin{gathered} x=\frac{b_1{c}_2-b_2{c}_1}{a_1{b}_2-a_2{b}_1},y=\frac{c_1{a}_2-c_2{a}_1}{a_1{b}_2-a_2{b}_1} \\ x=\frac{bx-(a+b)-(-a)x-(a-b)}{ax-a-bxb}=\frac{-ab-b^2-a^2+ab}{-a^2-b^2}=\frac{-a^2-b^2}{-a^2-b^2}=1 \\ y=\frac{-(a-b)xb-(-(a+b\left)\right)xa}{ax-a-bxb}=\frac{-ab+b^2+a^2+ab}{-a^2-b^2}=\frac{a^2+b^2}{-(a^2+b^2)}=-1 \end{gathered}$$

Hence x=1 & y=-1 is the required solution