Question

Solve the following simultaneous equation using Cramer’s rule.
$3x-4y=10$ and $4x+3y=5$

Answer 1

We are given a question with simultaneous equations which we have to solve using the Cramer’s rule. As per Cramer’s rule, $x=\dfrac{{{\Delta }_{x}}}{\Delta }$ and $y=\dfrac{{{\Delta }_{y}}}{\Delta }$, where $\Delta$ is the discriminant and it should not be equal to 0. From the two equations given to us, we will first find the corresponding values to be put in the discriminant. We will find the required discriminant and substitute the value in the Cramer’s equations and hence, we will have the values of ‘x’ and ‘y’.

Complete step-by-step answer:
According to the given question, we are given a set of equations which we have to solve using the Cramer’s rule.
The given equation is in the form $ax+by=c$. So the two equations can be expressed as,
$3x-4y=10$ can be expressed as ${{a}_{1}}x+{{b}_{1}}y={{c}_{1}}$ with the corresponding values and
$4x+3y=5$ can be expressed as ${{a}_{2}}x+{{b}_{2}}y={{c}_{2}}$ with the corresponding values as well.
As per Cramer’s rule, in order to find the value of ‘x’ and ‘y’, we have,
$x=\dfrac{{{\Delta }_{x}}}{\Delta }$ and $y=\dfrac{{{\Delta }_{y}}}{\Delta }$, where $\Delta$ is the discriminant and it should not be equal to 0.
Here, $\Delta =\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} \\\ {{a}_{2}} & {{b}_{2}} \\\ \end{matrix} \right|$, ${{\Delta }_{x}}=\left| \begin{matrix} {{c}_{1}} & {{b}_{1}} \\\ {{c}_{2}} & {{b}_{2}} \\\ \end{matrix} \right|$ and ${{\Delta }_{y}}=\left| \begin{matrix} {{a}_{1}} & {{c}_{1}} \\\ {{a}_{2}} & {{c}_{2}} \\\ \end{matrix} \right|$
We will now substitute the value in the above determinants and find the values. We get,

{{a}_{1}} & {{b}_{1}} \\\ {{a}_{2}} & {{b}_{2}} \\\ \end{matrix} \right|=\left| \begin{matrix} 3 & -4 \\\ 4 & 3 \\\ \end{matrix} \right|=\left( 3\times 3 \right)-\left( -4\times 4 \right)$$ $$\Delta =9-\left( -16 \right)$$ Adding up, we get the value as, $$\Delta =9+16=25\ne 0$$ Next, we have, $${{\Delta }_{x}}=\left| \begin{matrix} {{c}_{1}} & {{b}_{1}} \\\ {{c}_{2}} & {{b}_{2}} \\\ \end{matrix} \right|=\left| \begin{matrix} 10 & -4 \\\ 5 & 3 \\\ \end{matrix} \right|=\left( 10\times 3 \right)-\left( -4\times 5 \right)$$ $${{\Delta }_{x}}=30-\left( -20 \right)$$ Again, adding up the terms, we get, $${{\Delta }_{x}}=30+20=50$$ And next, we have, $${{\Delta }_{y}}=\left| \begin{matrix} {{a}_{1}} & {{c}_{1}} \\\ {{a}_{2}} & {{c}_{2}} \\\ \end{matrix} \right|=\left| \begin{matrix} 3 & 10 \\\ 4 & 5 \\\ \end{matrix} \right|=\left( 3\times 5 \right)-\left( 10\times 4 \right)$$ $${{\Delta }_{y}}=15-40=-25$$ Cramer’s equations that we have, $$x=\dfrac{{{\Delta }_{x}}}{\Delta }$$ and $$y=\dfrac{{{\Delta }_{y}}}{\Delta }$$ Substituting the values in the above equations, we get, $$x=\dfrac{50}{25}=2$$ And $$y=\dfrac{-25}{25}=-1$$ Therefore, the value of $$x=2,y=-1$$. **Note:** The discriminant should not be zero, because then the Cramer’s equations will give us an undefined value of x and y. Also, the value of the determinant should be calculated step – wise to prevent any mistakes. And while substituting the values in the Cramer’s equations, the values should not get interchanged.