# Solve the following system of equations by cross-multiplication method ax + by = 1; - Mathematics

Sum

Solve the following system of equations by cross-multiplication method ax + by = 1;  bx + ay = \frac{(a+b)^{2}}{a^{2}+b^{2}-1

#### Solution

The given system of equations can be written as

ax + by – 1 = 0 ….(1)

bx+ay=\frac{(a+b)^{2}}{a^{2}+b^{2}}-1

\Rightarrowbx+ay=\frac{a^{2}+2ab+b^{2}-a^{2}-b^{2}}{a^{2}+b^{2}}

\Rightarrow bx+ay=\frac{2ab}{a^{2}+b^{2}}

\Rightarrow bx+ay-\frac{2ab}{a^{2}+b^{2}}=0 ….. (2)

Rewritting the equations (1) and (2), we have

ax + by – 1 = 0

\Rightarrow bx+ay-\frac{2ab}{a^{2}+b^{2}}=0

Now, by cross-multiplication method, we have

\Rightarrow \frac{x}{b\times( \frac{-2ab}{a^{2}+b^{2}})-a\times (-1)}=\frac{-y}{a\times(\frac{-2ab}{a^{2}+b^{2}))-b\times (-1)}=\frac{1}{a\timesa-b\times b}

\Rightarrow\frac{x}{-\frac{2ab^{2}}{a^{2}+b^{2}}+a}=-y/((-2a^2b)/(a^2+b^2)+ b)=\frac{1}{a^{2}-b^2}

\Rightarrow\frac{x}{\frac{-2ab^{2}+a^{3}+ab^{2}}{a^{2}+b^{2}}}=-y/((-2a^2b+a^2b+b^3)/(a^2+b^2))=\frac{1}{a^{2}-b^{2}

=>x/((a(a^2-b^2))/(a^2+b^2))=(-y)/((b(b^2-a^2))/(a^2+b^2))=1/(a^2-b^2)

=>x/((a(a^2-b^2))/(a^2+b^2))=1/(a^2+b^2)=>x=a/(a^2+b^2)

"and "(-y)/((b(b^2-a^2))/(a^2+b^2))=1/(a^2+b^2)=>y=b/(a^2+b^2)

Hence, the solution is  x=\frac{a}{a^{2}+b^{2}},y=\frac{b}{a^{2}+b^{2}}

Concept: Algebraic Methods of Solving a Pair of Linear Equations - Cross - Multiplication Method
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