Question

Given that both x and y are positive integers. Solve the following systems of equations.
\[{{x}^{2}}+xy=15\]
\[{{y}^{2}}+xy=10\]

Answer 1

Hint: To solve the given question, we will first add both the equations given in the question. After adding these two equations, we will get the value of (x + y). Then, we will take x common from the left-hand side of the first equation and cross multiply x. Then we will square this equation obtained and put the value of x + y from the previous equation into this equation. From this equation, we will get the value of x. Then we will put this value of x in the second equation which will give us a quadratic equation in y. Then, we will solve this quadratic equation by using the quadratic formula. The positive value of y will be the required answer.

Complete step-by-step answer:
To start with, we are given the following two equations:
\[{{x}^{2}}+xy=15.......\left( i \right)\]
\[{{y}^{2}}+xy=10.........\left( ii \right)\]
Now, we will add both these equations. Thus, we will get,
\[\Rightarrow {{x}^{2}}+xy+{{y}^{2}}+xy=15+10\]
\[\Rightarrow {{x}^{2}}+2xy+{{y}^{2}}=25\]
Now, here, we will use the identity \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}.\] Thus, we will get,
\[\Rightarrow {{\left( x+y \right)}^{2}}=25......\left( iii \right)\]
Now, from equation (i), we have,
\[{{x}^{2}}+xy=15\]
Taking x common from the above equation, we will get,
\[\Rightarrow x\left( x+y \right)=15\]
Now, we will cross multiply x. Thus, we will get,
\[\Rightarrow \left( x+y \right)=\dfrac{15}{x}\]
Now, we will square both sides of the equation. Thus, we will get,
\[\Rightarrow {{\left( x+y \right)}^{2}}={{\left( \dfrac{15}{x} \right)}^{2}}........\left( iv \right)\]
From (iii) and (iv), we will get,
\[\Rightarrow 25={{\left( \dfrac{15}{x} \right)}^{2}}\]
Taking square root on both the sides, we will get,
\[\Rightarrow \sqrt{25}=\sqrt{{{\left( \dfrac{15}{x} \right)}^{2}}}\]
\[\Rightarrow 5=\dfrac{15}{x}\]
\[\Rightarrow x=3......\left( v \right)\]
Now, we will put the value of x = 3 in (ii). Thus, we will get,
\[{{y}^{2}}+3y=10\]
\[\Rightarrow {{y}^{2}}+3y-10=0\]
Now, we will solve this quadratic equation in y using the quadratic formula. If there is a quadratic equation of the form \[a{{x}^{2}}+bx+c=0\] then its roots are given by \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}.\]
Thus, we can say that,
\[y=\dfrac{-3\pm \sqrt{{{\left( 3 \right)}^{2}}-4\left( 1 \right)\left( -10 \right)}}{2\left( 1 \right)}\]
\[\Rightarrow y=\dfrac{-3\pm \sqrt{9+40}}{2}\]
\[\Rightarrow y=\dfrac{-3\pm \sqrt{49}}{2}\]
\[\Rightarrow y=\dfrac{-3\pm 7}{2}\]
\[y=\dfrac{-3+7}{2}\text{ or }y=\dfrac{-3-7}{2}\]
\[\Rightarrow y=\dfrac{4}{2}\text{ or }y=\dfrac{-10}{2}\]
\[\Rightarrow y=2\text{ or }y=-5\]
Now, y is positive, so y = 2. Thus, the value of x is 3 and y is 2.

Note: The alternate way of solving the question is shown below. From (i), we have,
\[{{x}^{2}}+xy=15\]
\[\Rightarrow x\left( x+y \right)=15\]
\[\Rightarrow x+y=\dfrac{15}{x}\]
\[\Rightarrow y=\dfrac{15}{x}-x\]
\[\Rightarrow y=\dfrac{15-{{x}^{2}}}{x}\]
Now, we will put this value of y in (ii). This, we will get,
\[{{\left( \dfrac{15-{{x}^{2}}}{x} \right)}^{2}}+x\left( \dfrac{15-{{x}^{2}}}{x} \right)=10\]
\[\Rightarrow \dfrac{225+{{x}^{4}}-30{{x}^{2}}}{{{x}^{2}}}+15-{{x}^{2}}=10\]
\[\Rightarrow \dfrac{225+{{x}^{4}}-30{{x}^{2}}}{{{x}^{2}}}={{x}^{2}}-5\]
\[\Rightarrow 225+{{x}^{4}}-30{{x}^{2}}={{x}^{2}}\left( {{x}^{2}}-5 \right)\]
\[\Rightarrow 225+{{x}^{4}}-30{{x}^{2}}={{x}^{4}}-5{{x}^{2}}\]
\[\Rightarrow 225-30{{x}^{2}}=-5{{x}^{2}}\]
\[\Rightarrow 225=25{{x}^{2}}\]
\[\Rightarrow {{x}^{2}}=9\]
\[\Rightarrow x=3\]
On putting this value of x in (i), we will get,
\[\Rightarrow {{\left( 3 \right)}^{2}}+3\left( y \right)=15\]
\[\Rightarrow 3y+9=15\]
\[\Rightarrow 3y=6\]
\[\Rightarrow y=2\]