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
380.7k+ views
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., $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
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., $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
Recently Updated Pages
Define absolute refractive index of a medium

Find out what do the algal bloom and redtides sign class 10 biology CBSE

Prove that the function fleft x right xn is continuous class 12 maths CBSE

Find the values of other five trigonometric functions class 10 maths CBSE

Find the values of other five trigonometric ratios class 10 maths CBSE

Find the values of other five trigonometric functions class 10 maths CBSE

Trending doubts
State Gay Lusaaccs law of gaseous volume class 11 chemistry CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

What is pollution? How many types of pollution? Define it

Change the following sentences into negative and interrogative class 10 english CBSE

Which is the tallest animal on the earth A Giraffes class 9 social science CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

How fast is 60 miles per hour in kilometres per ho class 10 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
