Question

Solve the following equations:
\[2{{x}^{2}}-xy+{{y}^{2}}=2y\] and \[2{{x}^{2}}+4xy=5y\].
a) \[\left( 0,0 \right)\]
b) \[\left( 1,2 \right)\]
c) \[\left( 2,3 \right)\]
d) \[\left( 0,1 \right)\]

Answer 1

Hint: In this question, we can use elimination methods to find the value of one variable in terms of other variable, and then we can apply substitution. In such a manner we will get the values of x and y. We can see that we require one pair of x and y, but as the equations are quadratic, we will get 2 pairs of values of x and y. So, we will use options to find the correct solution.

Complete step by step answer:
In this question, we are given with a pair of equations \[2{{x}^{2}}-xy+{{y}^{2}}=2y\] and \[2{{x}^{2}}+4xy=5y\], and we have to find the values of x and y, which satisfies the condition of both the equation.
Let us assume, \[2{{x}^{2}}-xy+{{y}^{2}}=2y......\left( i \right)\] and \[2{{x}^{2}}+4xy=5y......\left( ii \right)\]. Now, to find the values of x and y, we will use the most basic method of solving pairs of equations, that is, elimination method.
To apply elimination method, we require coefficients of at least one variable to be the same in both the equations. And, we can see, in both the equations, the coefficients of \[{{x}^{2}}\] are equal. So, we will subtract equation (i) from equation (ii), we get,
 \[2{{x}^{2}}+4xy-2{{x}^{2}}+xy-{{y}^{2}}=5y-2y\]
On further simplifying, we will get,
\[\Rightarrow 5xy-{{y}^{2}}=3y\]
Here, we can express x in terms of y
\[x=\dfrac{3y+{{y}^{2}}}{5y}\]
\[\Rightarrow x=\dfrac{3+y}{5}......\left( iii \right)\]
Now, we will put the value of x in terms of y from equation (iii) to equation (i), which implies
\[2{{\left( \dfrac{3+y}{5} \right)}^{2}}-\left( \dfrac{3+y}{5} \right)y+{{y}^{2}}=2y\]
\[\Rightarrow 2\left( \dfrac{9+6y+{{y}^{2}}}{25} \right)-\left( \dfrac{3+y}{5} \right)y+{{y}^{2}}=2y\]
We can further simplify it as,
\[\left( \dfrac{18+12y+2{{y}^{2}}}{25} \right)-\left( \dfrac{3+y}{5} \right)y+{{y}^{2}}=2y\]
Now, we will take L.C.M. on L.H.S., so we get,
\[\dfrac{18+12y+2{{y}^{2}}-15y-5{{y}^{2}}+25{{y}^{2}}}{25}=2y\]
Now, we will apply cross multiplication on both sides, we will get,
\[18+12y+2{{y}^{2}}-15y-5{{y}^{2}}+25{{y}^{2}}=50y\]
\[\Rightarrow 22{{y}^{2}}-53y+18=0\]
Now, we will apply Sridharacharya’s formula, to find the value of y from the obtained quadratic equation, which states that for a quadratic equation \[a{{y}^{2}}+by+c=0\], the roots are equal to \[y=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\], so the roots of quadratic equation \[22{{y}^{2}}-53y+18=0\] are
\[y=\dfrac{-\left( -53 \right)\pm \sqrt{{{\left( -53 \right)}^{2}}-4\left( 22 \right)\left( 18 \right)}}{2\left( 22 \right)}\]
\[\Rightarrow y=\dfrac{53\pm \sqrt{2809-1584}}{44}\]
\[\Rightarrow y=\dfrac{53\pm \sqrt{1225}}{44}\]
\[\Rightarrow y=\dfrac{53\pm 35}{44}\]
\[\Rightarrow y=\dfrac{88}{44},\dfrac{18}{44}\]
\[\Rightarrow y=2,\dfrac{9}{22}\]
From options, we can see that x and y have only the whole number as their solutions, so the only possible value of y can be 2.
Now, we will put value of y in equation (iii), we will get,
\[x=\dfrac{3+2}{5}\]
\[\Rightarrow x=\dfrac{5}{5}\]
\[\Rightarrow x=1\]
Therefore, the value of x is 1.
Hence, the solution of the given pair of equations is (1,2), that means, option (b) is correct. But if we put the values of x and y option wise then (0,0) also satisfies the given pair of equations. So, option (a) and option (b) both satisfy the pair of equations.

Note: Here, in this question, we can also apply a substitution method, to find the values of x and y. We can also say that we can find the value of one variable in terms of another variable from equation (ii) and can substitute that in equation (i). And, we will get the values of y by applying Sridharacharya’s formula. And, then we can calculate the value of x by substituting the value of y in any of the equation (i) or (ii).