Question

Find the values of x and y using cross – multiplication method: 3x + 4y = 43 and – 2x + 3y = 11.
(a) (7, 2)
(b) (– 7, 7)
(c) (5, 7)
(d) (– 6, – 7)

Answer 1

Hint: First take all the terms in the given equations to the LHS and then assume the two equations as equation (i) and equation (ii). Now, assume that the coefficients of x, y and constant term of the equation (i) are \[{{a}_{1}},{{b}_{1}},{{c}_{1}}\] respectively and the coefficients of x, y and constant term of equation (ii) are \[{{a}_{2}},{{b}_{2}},{{c}_{2}}\] respectively. Now, apply the formula for the 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 solution:
Here we have been provided with two equations:
\[3x+4y=43\]
\[-2x+3y=11\]
Taking all the terms to the LHS, we get,
\[3x+4y-43=0.......\left( i \right)\]
\[-2x+3y-11=0.......\left( ii \right)\]
Assuming the coefficients of x, y and constant terms of the equation (i) as \[{{a}_{1}},{{b}_{1}},{{c}_{1}}\] respectively and coefficients of x, y and constant term of equation (ii) as \[{{a}_{2}},{{b}_{2}},{{c}_{2}}\] respectively, we have,
\[{{a}_{1}}=3,{{b}_{1}}=4,{{c}_{1}}=-43\]
\[{{a}_{2}}=-2,{{b}_{2}}=3,{{c}_{2}}=-11\]
Applying the formula of cross – multiplication method, we have,
\[\dfrac{x}{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}=\dfrac{y}{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}=\dfrac{1}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\]
Therefore, substituting all the values of the coefficients, we have,
\[\Rightarrow \dfrac{x}{4\times \left( -11 \right)-3\times \left( -43 \right)}=\dfrac{y}{\left( -43 \right)\times \left( -2 \right)-\left( -11 \right)\times 3}=\dfrac{1}{3\times 3-\left( -2 \right)\times 4}\]
\[\Rightarrow \dfrac{x}{-44+129}=\dfrac{y}{86+33}=\dfrac{1}{9+8}\]
\[\Rightarrow \dfrac{x}{85}=\dfrac{y}{119}=\dfrac{1}{17}\]
Equating \[\dfrac{x}{85}=\dfrac{1}{17},\] we get,
\[\Rightarrow x=\dfrac{85}{17}\]
On simplifying, we get,
\[\Rightarrow x=5\]
Now, equating \[\dfrac{y}{119}=\dfrac{1}{17},\] we get,
\[\Rightarrow y=\dfrac{119}{17}\]
On simplifying, we get,
\[\Rightarrow y=7\]
Hence, the solution of the given system of linear equations is (x, y) = (5, 7).
Therefore, option (c) is the right answer.

Note: One may note that we can also solve this question with the help of substitution or elimination method but it is 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. Anyone who signs mistakes 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.