Question

Based on the cross-multiplication method, solve the following pairs of the equation by cross-multiplication rule.
\[\dfrac{x}{2} - \dfrac{y}{3} + 4 = 0,{\text{ }}\dfrac{x}{2} - \dfrac{{5y}}{3} + 12 = 0\]
Then x + y is equal to

Answer 1

Hint: We have to only use cross multiplication condition which states that if \[{a_1}x + {b_1}y + {c_1} = 0\] and \[{a_2}x + {b_2}y + {c_2} = 0\] are the two linear equations then it can also be written as \[\dfrac{x}{{\left( {{b_1}{c_2} - {b_2}{c_1}} \right)}} = \dfrac{y}{{\left( {{c_1}{a_2} - {c_2}{a_1}} \right)}} = \dfrac{1}{{\left( {{a_1}{b_2} - {a_2}{b_1}} \right)}}\].

Complete step-by-step answer:

To solve the given system of linear equations by cross-multiplication method. We have to first, write them in form of \[ax + by + c = 0\]
Now as we know that we are given with two linear equations and that were,
\[\dfrac{x}{2} - \dfrac{y}{3} + 4 = 0\] $\to$ (1)
\[\dfrac{x}{2} - \dfrac{{5y}}{3} + 12 = 0\] $\to$ (2)
So, we had to write equation 1 and 2 in the form of \[ax + by + c = 0\].
So, taking LCM in LHS of equation 1 and equation 2. And then cross multiplying both sides of the equation. We get,
3x – 2y + 24 = 0 $\to$ (3)
3x – 10y + 72 = 0 $\to$ (4)
Now as we know that if \[{a_1}x + {b_1}y + {c_1} = 0\] and \[{a_2}x + {b_2}y + {c_2} = 0\] are the two linear equations then by cross multiplication method. We can write,
\[\dfrac{x}{{\left( {{b_1}{c_2} - {b_2}{c_1}} \right)}} = \dfrac{y}{{\left( {{c_1}{a_2} - {c_2}{a_1}} \right)}} = \dfrac{1}{{\left( {{a_1}{b_2} - {a_2}{b_1}} \right)}}\]
So, \[x = \dfrac{{\left( {{b_1}{c_2} - {b_2}{c_1}} \right)}}{{\left( {{a_1}{b_2} - {a_2}{b_1}} \right)}}\] and \[y = \dfrac{{\left( {{c_1}{a_2} - {c_2}{a_1}} \right)}}{{\left( {{a_1}{b_2} - {a_2}{b_1}} \right)}}\]
So, let us apply the above formula of the two in the equation to find the value of x and y. We get,
\[x = \dfrac{{\left( {{b_1}{c_2} - {b_2}{c_1}} \right)}}{{\left( {{a_1}{b_2} - {a_2}{b_1}} \right)}} = \dfrac{{\left( { - 2 \times 72 - \left( { - 10} \right) \times 24} \right)}}{{\left( {3 \times \left( { - 10} \right) - 3 \times \left( { - 2} \right)} \right)}} = \dfrac{{96}}{{ - 24}} = - 4\]
And, \[y = \dfrac{{\left( {{c_1}{a_2} - {c_2}{a_1}} \right)}}{{\left( {{a_1}{b_2} - {a_2}{b_1}} \right)}} = \dfrac{{\left( {24 \times 3 - 72 \times 3} \right)}}{{\left( {3 \times \left( { - 10} \right) - 3 \times \left( { - 2} \right)} \right)}} = \dfrac{{ - 144}}{{ - 24}} = 6\]
Now x = - 4 and y = 6. So, x + y = 2.

Note: Whenever we come up with this type of question where we are asked to find the solution of given system of equations by using cross-multiplication method then we first write the given equations in form of \[ax + by + c = 0\] and after that we can compare the given equations with \[{a_1}x + {b_1}y + {c_1} = 0\] and \[{a_2}x + {b_2}y + {c_2} = 0\]. And apply the direct formula to find the value of x and y using cross-multiplication method which is \[x = \dfrac{{\left( {{b_1}{c_2} - {b_2}{c_1}} \right)}}{{\left( {{a_1}{b_2} - {a_2}{b_1}} \right)}}\] and \[y = \dfrac{{\left( {{c_1}{a_2} - {c_2}{a_1}} \right)}}{{\left( {{a_1}{b_2} - {a_2}{b_1}} \right)}}\].