Solve the equations using the cross multiplication method: \[3x + 2y = 10\] and \[4x - 2y = 4\] .
A.x=-2, y=2
B.x=2, y=-2
C.x=-2, y=-2
D.x=2, y=2

Hint: We use the cross multiplication method to solve linear equations in two variables. If \[{a_1}x + {b_1}y = {c_1}\] and \[{a_2}x + {b_2}y = {c_2}\] two linear equation, then the cross multiplication method we have, \[\dfrac{x}{{({b_1}{c_2} - {b_2}{c_1})}} = \dfrac{y}{{({c_1}{a_2} - {a_1}{c_2})}} = \dfrac{{ - 1}}{{({a_1}{b_2} - {a_2}{b_1})}}\] . Simplifying this we will get the roots for the two linear equations.

Complete step-by-step answer:
We have two linear equations, \[3x + 2y = 10\] and \[4x - 2y = 4\] .
Comparing these two linear equation with \[{a_1}x + {b_1}y = {c_1}\] and \[{a_2}x + {b_2}y = {c_2}\] , we have \[{a_1} = 3\] , \[{b_1} = 2\] , \[{c_1} = 10\] , \[{a_2} = 4\] , \[{b_2} = - 2\] and \[{c_2} = 4\] .
As by the cross multiplication method we have \[\dfrac{x}{{({b_1}{c_2} - {b_2}{c_1})}} = \dfrac{y}{{({c_1}{a_2} - {a_1}{c_2})}} = \dfrac{{ - 1}}{{({a_1}{b_2} - {a_2}{b_1})}}\] .
Substituting we have,
 \[ \Rightarrow \dfrac{x}{{\left( {(2 \times 4) - ( - 2 \times 10)} \right)}} = \dfrac{y}{{\left( {(10 \times 4) - (3 \times 4)} \right)}} = \dfrac{{ - 1}}{{\left( {(3 \times - 2) - (4 \times 2)} \right)}}\]
Simplifying we get,
 \[ \Rightarrow \dfrac{x}{{\left( {8 - ( - 20)} \right)}} = \dfrac{y}{{\left( {40 - 12} \right)}} = \dfrac{{ - 1}}{{\left( {( - 6) - 8} \right)}}\]
Negative multiplied by negative is positive.
 \[ \Rightarrow \dfrac{x}{{\left( {8 + 20} \right)}} = \dfrac{y}{{\left( {40 - 12} \right)}} = \dfrac{{ - 1}}{{\left( { - 6 - 8} \right)}}\]
 \[ \Rightarrow \dfrac{x}{{28}} = \dfrac{y}{{28}} = \dfrac{{ - 1}}{{ - 14}}\]
 \[ \Rightarrow \dfrac{x}{{28}} = \dfrac{y}{{28}} = \dfrac{1}{{14}}\]
Now, we have
 \[ \Rightarrow \dfrac{x}{{28}} = \dfrac{1}{{14}}\] and \[ \Rightarrow \dfrac{y}{{28}} = \dfrac{1}{{14}}\] .
Multiplied by 28 on both sides, in both terms.
 \[ \Rightarrow x = \dfrac{{28}}{{14}}\] and \[ \Rightarrow y = \dfrac{{28}}{{14}}\]
Simple division, we have
 \[ \Rightarrow x = 2\] and \[ \Rightarrow y = 2\] .
That is we have x=2, y=2.
So, the correct answer is “x=2, y=2”.

Note: Careful about substituting the coefficients of x and y. The procedure is the same for any linear equation with two variables. Follow the same procedure as above. We can also solve this problem by balancing coefficients of any one of the variables. Balance the coefficient of x by multiplying coefficient of given equation with each other (like coefficient of first equation multiplied with whole second equation and coefficient of second equation multiplied with whole first equation). Afterwards we can cancel one of the variables and we get another variable. After substituting we can find the other variable as well. In both the cases we will have the same answer.