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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) $12,5$

Answer Verified Verified
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.

Complete step-by-step answer:
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., $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
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