Question

Solve using cross-multiplication:
$3x+4y=11$
$2x+3y=8$

Answer 1

Hint: First of all take all the terms in the given equations to the L.H.S and then assume the two equations as equation (i) and equation (ii). Now, assume the coefficients of x, y and the constant term of equation (i) as ${{a}_{1}},{{b}_{1}}$ and ${{c}_{1}}$ respectively and the coefficients of x, y and the constant term of equation (ii) as ${{a}_{2}},{{b}_{2}}$ and ${{c}_{2}}$ respectively. Now, apply the formula for method of cross-multiplication: \[\dfrac{x}{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}=\dfrac{y}{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}=\dfrac{1}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\] and equate \[\dfrac{x}{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}=\dfrac{1}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\] to find the value of x and \[\dfrac{y}{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}=\dfrac{1}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\] to find the value of y.

Complete step-by-step answer:
Here we have been provided with two equations $3x+4y=11$ and $2x+3y=8$ and we are asked to solve this system using the cross multiplication method. First let use simplify the two equations by taking all the terms to the L.H.S, so we get,
$\Rightarrow 3x+4y-11=0.........\left( i \right)$
$\Rightarrow 2x+3y-8=0.........\left( ii \right)$
Assuming the coefficients of x, y and the constant term of equation (i) as ${{a}_{1}},{{b}_{1}}$ and ${{c}_{1}}$ respectively and the coefficients of x, y and the constant term of equation (ii) as ${{a}_{2}},{{b}_{2}}$ and ${{c}_{2}}$ respectively, we have,
$\Rightarrow {{a}_{1}}=3,{{b}_{1}}=4$ and ${{c}_{1}}=-11$
$\Rightarrow {{a}_{2}}=2,{{b}_{2}}=3$ and ${{c}_{2}}=-8$
Applying the formula of cross-multiplication method we have,
\[\Rightarrow \dfrac{x}{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}=\dfrac{y}{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}=\dfrac{1}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\]
Therefore substituting all the values of coefficients and constants we get,
\[\begin{align}
  & \Rightarrow \dfrac{x}{\left( 4 \right)\left( -8 \right)-\left( 3 \right)\left( -11 \right)}=\dfrac{y}{\left( -11 \right)\left( 2 \right)-\left( -8 \right)\left( 3 \right)}=\dfrac{1}{\left( 3 \right)\left( 3 \right)-\left( 2 \right)\left( 4 \right)} \\
 & \Rightarrow \dfrac{x}{-32+33}=\dfrac{y}{-22+24}=\dfrac{1}{9-8} \\
 & \Rightarrow \dfrac{x}{1}=\dfrac{y}{2}=\dfrac{1}{1} \\
\end{align}\]
(i) For the value of x we have,
 \[\begin{align}
  & \Rightarrow \dfrac{x}{1}=\dfrac{1}{1} \\
 & \Rightarrow x=1 \\
\end{align}\]
(ii) For the value of y we have,
\[\begin{align}
  & \Rightarrow \dfrac{y}{2}=\dfrac{1}{1} \\
 & \Rightarrow y=2 \\
\end{align}\]
Hence, the solution of the given system of equations is (1, 2).

Note: Note that we can also solve this question easily by the help of substitution or elimination method but it its asked in the question to solve it with the help of cross-multiplication method. While solving this question, we have to be careful about the formula and calculation. Any one sign mistake can lead us to the wrong answer. We can check our answer by substituting the obtained values of x and y in the provided equations.