Question

Solve the following pairs of equations.
$43x+67y=-24$ and $67x+43y=24$

Answer 1

Hint: Use elimination method, any of the coefficients x or y is first equated and eliminated. After elimination the equations are solved to obtain the other equation. Thus, you get the value x and y.

Complete step-by-step answer:
We have been given two linear equations which we have to solve and get the value of x and y. The two linear equations are respectively,
$43x+67y=-24$ ………………………………(1)
$67x+43y=24$ ………………………………….(2)
We can solve the linear equation by using the Elimination method, any of the coefficients is first equated and eliminated. After elimination, the equations are solved to obtain the other equation.
Let us consider equation (1), multiply the entire equation by 67. Now in equation (2), multiply the entire equation by 43.
$43x+67y=-24$ ………………………………(1) $\times 67$
$67x+43y=24$ ………………………………….(2) $\times 43$
Thus, we get the new equations as,
$\begin{align}
  & 2881x+4489y=-1608 \\
 & \underline{2881x+1849y=\text{ 1032}} \\
\end{align}$
Now let us subtract both the equations,
\[\begin{align}
  & \text{ }2881x+4489y=-1608 \\
 & \underline{{}^{\left( - \right)}2881x\overset{\left( - \right)}{\mathop{+}}\,1849y={}^{\left( - \right)}\text{1032}} \\
 & \text{ 2640}y=-2640 \\
\end{align}\]
Thus, we get \[y=\dfrac{-2640}{2640}=-1\]
Hence, we got the value of \[y=-1\].
Now let us put the \[y=-1\] in equation (2)
$67x+43y=24$
$67x+43\times \left( -1 \right)=24$
\[\therefore 67x=24+43\]
\[x=\dfrac{67}{67}=1\]
\[\therefore \] we got the value of x and y as 1 and \[-1\] .
\[\therefore x=1\] and \[y=-1\]

Note: We can solve this using a substitution method.
From (1), $43x+67y=-24$
\[x=\dfrac{-\left( 24+67y \right)}{43}\] , put this in equation (2)
\[-67\left[ \dfrac{24+67y}{43} \right]+43y=24\]
\[\Rightarrow -67\left( 24+67y \right)+1846y=1032\]
\[-1608-4489y+1846y=1032\]
\[\Rightarrow -2640y=2640\]
\[\therefore y=-1\]