Question

Solve the following pair of linear equations:
(i) $px+qy=p-q;qx-py=p+q$
(ii) $ax+by=c;bx+ay=1+c$
(iii) $152x-378y=-74;-378x+152y=-604$

Answer 1

Hint:We solve the first two equations using the method of elimination. With the help of multiplication, we take out one variable from those equations to get a linear equation of the other variable. We solve them to get the solutions of the equations. For the third equation, we opt for addition to getting a third linear equation of both variables. Then by comparing we eliminate one variable. Finally using one value of the known variable we get the other one.

Complete step by step answer:
The first equation gives us $px+qy=p-q;qx-py=p+q$.
Let $px+qy=p-q$ …(i) and $qx-py=p+q$ …(ii)
We multiply the equation (i) with p and the equation (ii) with q to get ${{p}^{2}}x+pqy={{p}^{2}}-pq$ and ${{q}^{2}}x-pqy=pq+{{q}^{2}}$. Now we add these two equations to get to the value of x.
\[\begin{align}
  & {{p}^{2}}x+pqy+{{q}^{2}}x-pqy={{p}^{2}}-pq+pq+{{q}^{2}} \\
 & \Rightarrow x\left( {{p}^{2}}+{{q}^{2}} \right)={{p}^{2}}+{{q}^{2}} \\
 & \Rightarrow x=\dfrac{{{p}^{2}}+{{q}^{2}}}{{{p}^{2}}+{{q}^{2}}}=1 \\
\end{align}\]
Putting the value of x we get value of y as $p\left( 1 \right)+qy=p-q\Rightarrow qy=-q\Rightarrow y=\dfrac{-q}{q}=-1$.
So, the solution of the set of equations are $x=1,y=-1$.
The second equation gives us $ax+by=c;bx+ay=1+c$.
Let $ax+by=c$ …(i) and $bx+ay=1+c$ …(ii)
We multiply the equation (i) with a and the equation (ii) with b to get ${{a}^{2}}x+aby=ac$ and ${{b}^{2}}x+aby=b+bc$. Now we subtract the second one from the first equation to get to the value of x.
$\begin{align}
  & {{a}^{2}}x+aby-\left( {{b}^{2}}x+aby \right)=ac-\left( b+bc \right) \\
 & \Rightarrow {{a}^{2}}x+aby-{{b}^{2}}x-aby=ac-b-bc \\
 & \Rightarrow x\left( {{a}^{2}}-{{b}^{2}} \right)=ac-b-bc \\
 & \Rightarrow x=\dfrac{ac-b-bc}{{{a}^{2}}-{{b}^{2}}} \\
\end{align}$
Putting the value of $x=\dfrac{ac-b-bc}{{{a}^{2}}-{{b}^{2}}}=\dfrac{c}{a+b}-\dfrac{b}{{{a}^{2}}-{{b}^{2}}}$ we get value of y as
$\begin{align}
  & a\left( \dfrac{ac-b-bc}{{{a}^{2}}-{{b}^{2}}} \right)+by=c \\
 & \Rightarrow by=c-a\left( \dfrac{ac-b-bc}{{{a}^{2}}-{{b}^{2}}} \right) \\
 & \Rightarrow y=\dfrac{c\left( {{a}^{2}}-{{b}^{2}} \right)-a\left( ac-b-bc \right)}{{{a}^{2}}-{{b}^{2}}}=\dfrac{c\left( {{a}^{2}}-{{b}^{2}} \right)-ac\left( a-b \right)+ab}{{{a}^{2}}-{{b}^{2}}} \\
 & \Rightarrow y=\dfrac{\left( a-b \right)\left( ac+bc-ac \right)+ab}{{{a}^{2}}-{{b}^{2}}}=\dfrac{bc}{a+b}+\dfrac{ab}{{{a}^{2}}-{{b}^{2}}} \\
\end{align}$
So, the solution of the set of equations are $x=\dfrac{c}{a+b}-\dfrac{b}{{{a}^{2}}-{{b}^{2}}},y=\dfrac{bc}{a+b}+\dfrac{ab}{{{a}^{2}}-{{b}^{2}}}$.
The third equation gives us $152x-378y=-74;-378x+152y=-604$.
Let $152x-378y=-74$ …(i) and $-378x+152y=-604$ …(ii)
We add these two equations to get the value of $\left( x+y \right)$.
$\begin{align}
  & 152x-378y-378x+152y=-74-604 \\
 & \Rightarrow 152\left( x+y \right)-378\left( x+y \right)=-678 \\
 & \Rightarrow -226\left( x+y \right)=-678 \\
 & \Rightarrow x+y=\dfrac{-678}{-226}=3 \\
\end{align}$
Now we multiply with 378 to get $x+y=3\Rightarrow 378x+378y=1134$ …(iii)
We again add equation (i) and (iii) to get the value of x.
$\begin{align}
  & 152x-378y+378x+378y=-74+1134 \\
 & \Rightarrow 530x=1060 \\
 & \Rightarrow x=\dfrac{1060}{530}=2 \\
\end{align}$
Putting the value of x in equation (iii) we get value of y as $x+y=3\Rightarrow 2+y=3\Rightarrow y=1$.
So, the solution of the set of equations are $x=2,y=1$.

Note:
We need to remember that we have two equations of two unknowns. So, we can apply any kind of theorem or formula to get the solutions. It doesn’t have to be a fixed particular one.