Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

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

Answer
VerifiedVerified
543.9k+ views
Hint: In the given question, we have been given that there is a pair of linear equations in two variables. These equations are to be solved by cross-multiplication method. After solving them by the cross-multiplication rule, we have to calculate the sum of the two variables. For solving this question, we must know the formula of the cross-multiplication rule, how to arrange the coefficients, variables, constants in the correct manner. To solve the pair of equations for \[x{\rm{ }}and{\rm{ }}y\] using cross-multiplication, we arrange the variables \[x{\rm{ }}and{\rm{ }}y\] and their coefficients \[a1,{\rm{ }}a2,{\rm{ }}b1,{\rm{ }}b2\], and the constants \[c1,{\rm{ }}c2\] as given in the formula and just simply put in the values and find the variables. Then, we just need to add them to get our final answer.

Complete step-by-step answer:
The given pair of equations are:
\[\dfrac{x}{2} - \dfrac{y}{3} + 4 = 0\], \[\dfrac{x}{2} - \dfrac{{5y}}{3} + 12 = 0\]
First, let us consider equation \[\dfrac{x}{2} - \dfrac{y}{3} + 4 = 0\]
Multiplying the two sides of the equation by the LCM of denominators in the \[x\] and \[y\] fraction, which is
\[LCM\left( {2,3} \right) = 6\], we get,
$\Rightarrow$ \[6\left( {\dfrac{x}{2} - \dfrac{y}{3} + 4} \right) = 0\]
$\Rightarrow$ \[3x - 2y + 24 = 0\]
Similarly, for the second equation,
$\Rightarrow$ \[6\left( {\dfrac{x}{2} - \dfrac{{5y}}{3} + 12} \right) = 0\]
$\Rightarrow$ \[3x - 10y + 72 = 0\]
Now, we are going to use the formula for solving the given system of equations through cross-multiplication:
$\Rightarrow$ \[x = \dfrac{{{b_1}{c_2} - {b_2}{c_1}}}{{{a_1}{b_2} - {a_2}{b_1}}}\] and \[y = \dfrac{{{c_1}{a_2} - {c_2}{a_1}}}{{{a_1}{b_2} - {a_2}{b_1}}}\]
Putting in the values, we get,
$\Rightarrow$ \[x = \dfrac{{\left( {2 \times 72} \right) - \left( {10 \times 24} \right)}}{{\left( {3 \times 10} \right) - \left( {3 \times - 2} \right)}} = \dfrac{{144 - 240}}{{30 - 6}} = - \dfrac{{96}}{{24}} = - 4\]
$\Rightarrow$ \[y = \dfrac{{\left( {24 \times 3} \right) - \left( {72 \times 3} \right)}}{{\left( {3 \times - 10} \right) - \left( {3 \times - 2} \right)}} = \dfrac{{72 - 216}}{{ - 30 + 6}} = \dfrac{{ - 144}}{{ - 24}} = 6\]
Thus, \[x + y = - 4 + 6 = 2\]

Note: In cross-multiplication, we need to keep care of the things (numbers, equations, expressions) being added or subtracted from each other. As this is the thing where the most error occurs. Here, we multiplied both of the equations by the LCM of the denominators so as to simplify the equations being used.