Question

# Solve the following equations:$4{{x}^{2}}+5y=6+20xy-25{{y}^{2}}+2x;\text{ }7x-11y=17$.A) $4,1$B) $\dfrac{107}{13},\dfrac{48}{13}$C) $5,2$D) $12,5$

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Hint: We will have to perform the substitution method to solve the above given equations, i.e., solving for either x or y from one of the equations and substituting it in another equation.

Here, we have two equations to solve and there are two variables in those equations. So, it is possible for us to find out the unique solutions for the variables, i.e.,
$4{{x}^{2}}+5y=6+20xy-25{{y}^{2}}+2x\ldots \text{ }\left( 1 \right)$
$7x-11y=17\ldots \text{ }\left( 2 \right)$
From equation (1), we have a 2-degree equation so we can say that for a 2-degree equation we get two values for each variable. And we can solve the given equations by substituting the values of one variable in other, i.e.,
From equation (2),
\begin{align} & \Rightarrow 7x-11y=17 \\ & \Rightarrow 7x=17+11y \\ \end{align}
$\Rightarrow x=\dfrac{17+11y}{7}\ldots \text{ }\left( 3 \right)$
Now we will have to substitute this value of x in equation (1) and from equation (1), we have
$\Rightarrow 4{{x}^{2}}+5y=6+20xy-25{{y}^{2}}+2x$
Thus, substituting value of x from equation (3) in the above equation, we get
\begin{align} & \Rightarrow 4{{x}^{2}}+5y=6+20xy-25{{y}^{2}}+2x \\ & \Rightarrow 4{{\left( \dfrac{17+11y}{7} \right)}^{2}}+5y=6+20\left( \dfrac{17+11y}{7} \right)y-25{{y}^{2}}+2\left( \dfrac{17+11y}{7} \right) \\ \end{align}
Now, we have to solve the above equation for y by transposing and cross-multiplications.
Using the formula of ${{(a+b)}^{2}}=({{a}^{2}}+{{b}^{2}}+2ab)$, we get
\begin{align} & \Rightarrow 4{{\left( \dfrac{17+11y}{7} \right)}^{2}}+5y=6+20\left( \dfrac{17+11y}{7} \right)y-25{{y}^{2}}+2\left( \dfrac{17+11y}{7} \right) \\ & \Rightarrow 4\left( \dfrac{{{\left( 17 \right)}^{2}}+{{\left( 11y \right)}^{2}}+2\left( 17 \right)\left( 11y \right)}{{{7}^{2}}} \right)+5y=6+20\left( \dfrac{17+11y}{7} \right)y-25{{y}^{2}}+2\left( \dfrac{17+11y}{7} \right) \\ \end{align}
Now, opening the above formula with substituted values, we get
$\Rightarrow 4\left( \dfrac{289+121{{y}^{2}}+374y}{49} \right)+5y=6+20\left( \dfrac{17+11y}{7} \right)y-25{{y}^{2}}+2\left( \dfrac{17+11y}{7} \right)$
Taking the LCM of the denominators, we get
$\Rightarrow \left( \dfrac{4\left( 289+121{{y}^{2}}+374y \right)}{49} \right)+5y=6+20\left( \dfrac{17+11y}{7} \right)y-25{{y}^{2}}+2\left( \dfrac{17+11y}{7} \right)$
$\Rightarrow \left( \dfrac{1156+484{{y}^{2}}+1496y+\left( 49 \right)\left( 5y \right)}{49} \right)=\dfrac{\left( 6\times 7 \right)+20\left( 17+11y \right)y-\left( 25\times 7 \right){{y}^{2}}+2\left( 17+11y \right)}{7}$
$\Rightarrow \left( \dfrac{1156+484{{y}^{2}}+1496y+245y}{49} \right)=\dfrac{42+340y+220{{y}^{2}}-175{{y}^{2}}+34+22y}{7}$
On cross-multiplying the denominator from left-hand side to right-hand side, we get
\begin{align} & \Rightarrow \left( 1156+484{{y}^{2}}+1741y \right)=49\left( \dfrac{76+362y+45{{y}^{2}}}{7} \right) \\ & \Rightarrow \left( 1156+484{{y}^{2}}+1741y \right)=7\left( 76+362y+45{{y}^{2}} \right) \\ & \Rightarrow \left( 1156+484{{y}^{2}}+1741y \right)=\left( 532+2534y+315{{y}^{2}} \right) \\ \end{align}
Now, on transposing similar terms from LHS to RHS, we get
\begin{align} & \Rightarrow \left( 1156+484{{y}^{2}}+1741y \right)=\left( 532+2534y+315{{y}^{2}} \right) \\ & \Rightarrow \left( 1156-532+484{{y}^{2}}-315{{y}^{2}}+1741y-2534y \right)=0 \\ & \Rightarrow 169{{y}^{2}}-793y+624=0 \\ \end{align}
Rearranging terms in above equation, we get
\begin{align} & \Rightarrow 169{{y}^{2}}-793y+624=0 \\ & \Rightarrow 13{{y}^{2}}-61y+48=0\ldots \text{ }\left( 4 \right) \\ \end{align}
We can solve the equation (4), by performing middle-term split method, i.e.,
\begin{align} & \Rightarrow 13{{y}^{2}}-61y+48=0 \\ & \Rightarrow 13{{y}^{2}}-13y-48y+48=0 \\ & \Rightarrow 13y\left( y-1 \right)-48\left( y-1 \right)=0 \\ & \Rightarrow \left( 13y-48 \right)\left( y-1 \right)=0 \\ & \Rightarrow y=\dfrac{48}{13};\text{ }y=1 \\ \end{align}
Now, we have two values of y. Substituting the above values of y back in equation (3), we get
For $y=\dfrac{48}{13}$,
\begin{align} & \Rightarrow x=\dfrac{17+11y}{7} \\ & \Rightarrow x=\dfrac{17+\left( 11\times \dfrac{48}{13} \right)}{7} \\ & \Rightarrow x=\dfrac{221+528}{91} \\ & \Rightarrow x=\dfrac{749}{91}=\dfrac{107}{13} \\ \end{align}
We get, $\left( x,y \right)=\left( \dfrac{107}{13},\dfrac{48}{13} \right)$.
Also, substituting the other value of y in equation (3), we get
For $y=1$,
\begin{align} & \Rightarrow x=\dfrac{17+11y}{7} \\ & \Rightarrow x=\dfrac{17+\left( 11\times 1 \right)}{7} \\ & \Rightarrow x=\dfrac{28}{7} \\ & \Rightarrow x=4 \\ \end{align}
Thus, we get $\left( x,y \right)=\left( 4,1 \right)$.
Hence, the following solutions to our equations are: $\left( 4,\text{ }1 \right)\text{; }\left( \dfrac{107}{13},\text{ }\dfrac{48}{13} \right)$, i.e., Option (A), (B) are correct.

Note: There is a shortcut method to solve the above equations instead of performing lengthy middle term split method, we can use direct formula for solving roots of quadratic equation, i.e., $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.