Question

If we have the equations \[\dfrac{{x + y - 8}}{2} = \dfrac{{x + 2y - 14}}{3} = \dfrac{{3x - y}}{4}\] , then the values of \[x\] and \[y\] is
\[\left( 1 \right)\] \[x = 1\] , \[y = 3\]
\[\left( 2 \right)\] \[x = 5\] , \[y = 2\]
\[\left( 3 \right)\] \[x = 3\] , \[y = 3\]
\[\left( 4 \right)\] \[x = 2\] , \[y = 6\]

Answer 1

Hint: We have to find the value of both the variables x and y from the given expression \[\dfrac{{x + y - 8}}{2} = \dfrac{{x + 2y - 14}}{3} = \dfrac{{3x - y}}{4}\] . We solve this question using the concept of solving the linear equations . We should also have the knowledge of the concept of solving the equation using the elimination method . We will solve this by taking two pairs at a time , first we will take the first two pairs and then simplify the relation such that we obtain a linear equation . Then we will take the other pair of the expressions and then simplify the relation such that we obtain another linear equation and then we will solve the two linear equations using the elimination method . And hence find the value of the variable \[x\] and \[y\] .

Complete step-by-step solution:
Given :
\[\dfrac{{x + y - 8}}{2} = \dfrac{{x + 2y - 14}}{3} = \dfrac{{3x - y}}{4}\]
Let us solve this question by making two cases for the formation of the two linear equations .
\[Case{\text{ }}1{\text{ }}:\]
Let us take the first pair of expression as :
\[\dfrac{{x + y - 8}}{2} = \dfrac{{x + 2y - 14}}{3}\]
Now , we will simplify this expression for a linear equation as :
Cross multiplying the terms , the expression can be written as :
\[3 \times \left( {x + y - 8} \right) = 2 \times \left( {x + 2y - 14} \right)\]
On simplifying , we can write the expression as :
\[3x + 3y - 24 = 2x + 4y - 28\]
\[x - y = - 4--\left( 1 \right)\]
\[Case{\text{ }}2{\text{ }}:\]
Let us take the second pair of expression as :
\[\dfrac{{x + 2y - 14}}{3} = \dfrac{{3x - y}}{4}\]
Now , we will simplify this expression for another linear equation as :
Cross multiplying the terms , the expression can be written as :
\[4 \times \left( {x + 2y - 14} \right) = 3 \times \left( {3x - y} \right)\]
On simplifying , we can write the expression as :
\[4x + 8y - 56 = 9x - 3y\]
\[5x - 11y = - 56--\left( 2 \right)\]
Now , we have obtained the two linear equations for the given expression . Now we will solve these two equations using the elimination method of solving the equations .

On multiplying the equation \[\left( 1 \right)\] by \[5\] , we get the linear equation as :
\[5 \times \left[ {x - y = - 4} \right]\]
\[5x - 5y = - 20--\left( 3 \right)\]
On subtracting equation \[\left( 3 \right)\] from equation \[\left( 2 \right)\] , we get the value of \[y\] as :
\[5x - 11y - \left( {5x - 5y} \right) = - 56 - \left( { - 20} \right)\]
\[ - 11y + 5y = - 56 + 20\]
On further simplifying , we get the value of \[y\] as :
\[ - 6y = - 36\]
\[y = 6\]
Now , for the value of \[x\] .
On multiplying the equation \[\left( 1 \right)\] by \[11\] , we get the linear equation as :
\[11 \times \left[ {x - y = - 4} \right]\]
\[11x - 11y = - 44--\left( 4 \right)\]
On subtracting equation \[\left( 4 \right)\] from equation \[\left( 2 \right)\] , we get the value of \[x\] as :
\[5x - 11y - \left( {11x - 11y} \right) = - 56 - \left( { - 44} \right)\]
\[5x - 11x = - 56 + 44\]
On further simplifying , we get the value of \[x\] as :
\[ - 6x = - 12\]
\[x = 2\]
Hence , we get the value of the variables \[x\] and \[y\] as \[2\] and \[6\] respectively .
Thus , the correct option is \[\left( 4 \right)\] .

Note: For the value of the functional expression , we can use any of the methods of solving the equation . We could also use the substitution method or the cross multiplication method to solve the value of the functional expression . But we used the elimination method as it is easier and less complicated then the other two methods..