Question

Solve for $x$ and $y$:
 \[4x - 3y = 8\]
\[6x - y = \dfrac{{29}}{3}\].

Answer 1

Hint: Here in this given equation is a system of linear equations. We have to find the unknown values that are \[x\] and \[y\] solving these systems of equations by using the elimination method. In the elimination method either we add or subtract the equations to find the unknown.

Complete step by step solution:
Let us consider the equation and we will name it as (1) and (2)
\[4x - 3y = 8\]-------(1)
\[6x - y = \dfrac{{29}}{3}\]-------(2)
Now we have to solve these two equations to find the unknown
Multiply (2) by 3 on both side, then we get
\[18x - 3y = 29\]----------(3)
On subtracting equation (1) from equation (3), we get
\[ \Rightarrow \,\,equation\left( 3 \right) - equation\left( 1 \right)\]
\[ \Rightarrow \,\,18x - 3y - \left( {4x - 3y} \right) = 29 - 8\]
\[ \Rightarrow \,\,18x - 3y - 4x + 3y = 29 - 8\]
Since the coefficients of \[y\] are same but it has alternate, so we cancel the \[y\] term and by simplification we have
\[ \Rightarrow \,\,14x = 21\]
Dividing both side by 14, we have
\[ \Rightarrow \,\,x = \dfrac{{21}}{{14}}\]
Divide both numerator and denominator by 7, then
\[ \Rightarrow \,\,x = \dfrac{3}{2}\]
We have found the value of \[x\] now we have to find the value of \[y\]. So, we will substitute the value \[x\]to any one of the equations (1) or (2). we will substitute the value of \[x\]to equation (1).
Therefore, we have \[4x - 3y = 8\]
\[ \Rightarrow \,\,4\left( {\dfrac{3}{2}} \right) - 3y = 8\]
\[ \Rightarrow \,\,2\left( 3 \right) - 3y = 8\]
\[ \Rightarrow \,\,6 - 3y = 8\]
Subtract 6 on both side, then
\[ \Rightarrow \,\,6 - 3y - 6 = 8 - 6\]
On simplification, we get
\[ \Rightarrow \,\, - 3y = 2\]
Divide the above equation by -3 we have
\[ \Rightarrow \,y = - \dfrac{2}{3}\]
Therefore, the unknown values \[x\] and \[y\] that is \[\dfrac{3}{2}\] and \[\left( { - \dfrac{2}{3}} \right)\] respectively.

Note:
> We can check whether these values are correct or not by substituting the unknown values in the given equations and we have to prove L.H.S is equal to R.H.S
Now we will substitute the value of \[x\] and \[y\] in equation (1) so we have
\[4x - 3y = 8\]
\[ \Rightarrow 4\left( {\dfrac{3}{2}} \right) - 3\left( { - \dfrac{2}{3}} \right) = 8\]
\[ \Rightarrow 2\left( 3 \right) + 2 = 8\]
\[ \Rightarrow 6 + 2 = 8\]
\[ \Rightarrow 8 = 8\]
\[ \Rightarrow LHS = RHS\]
Hence the values of the unknown that are \[x\] and \[y\] are correct values which satisfies the equation.
> In this type of question while eliminating the term we must be aware of the sign where we change the sign by the alternate sign. In this we have a chance to verify our answers. In the elimination method we have made the one term have the same coefficient such that it will be easy to solve the equation.