Solve the following systems of equations.
$5\left( {x + y} \right) + 2xy = - 19,$$3xy + x + y = - 35$
Answer
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Hint: In the solution of any type of algebraic equation, firstly, we simply change the equation in a simple form by using the BODMAS rule. The BODMAS rule says that if the order of arithmetic operation should be according in this manner that first, we will open the bracket, then we will solve the orders, then division, multiplication, addition and subtraction will take place. So basically, BODMAS rules provide us the priority of operation in an equation when we simplify the equation.
Complete step-by-step answer:
Given equations are:
$5\left( {x + y} \right) + 2xy = - 19$
Apply BODMAS and solve the bracket first.
$\begin{array}{c}
5x + 5y + 2xy = - 19............{\rm{(1)}}\\
3xy + x + y = - 35............{\rm{(2)}}
\end{array}$
Multiply the equation (2) by $5$.
$\begin{array}{c}
5x + 5y + 2xy = - 19..............{\rm{(1)}}\\
15xy + 5x + 5y = - 175............{\rm{(3)}}
\end{array}$
Subtract equation (3) from equation (1).
$\begin{array}{c}
2xy - 15xy = - 19 - ( - 175)\\
- 13xy = 156\\
xy= - 12.................{\rm{(4)}}
\end{array}$
Substitute the value in equation (2) from equation (4)
$\begin{array}{c}
3\left( { - 12} \right) + x + y = - 35\\
x + y = 1\\
x = 1 - y...........(5)
\end{array}$
Substitute the value of $x$ in equation (4).
$\begin{array}{l}
\left( {1 - y} \right)y = - 12\\
{y^2} - y - 12 = 0
\end{array}$
$\begin{array}{l}
{y^2} - 4y + 3y - 12 = 0\\
y\left( {y - 4} \right) + 3\left( {y - 4} \right) = 0\\
\left( {y + 3} \right)\left( {y - 4} \right) = 0
\end{array}$
Evaluate further
$\begin{array}{l}
\left( {y + 3} \right) = 0,\left( {y - 4} \right) = 0\\
y = - 3,4..........{\rm{(6)}}
\end{array}$
Substitute the values of $y$ in equation (5).
$\begin{array}{l}
x = 1 - \left( { - 3} \right)\\
x = 4
\end{array}$
$\begin{array}{l}
x = 1 - 4\\
x = - 3
\end{array}$
So the answer is $\left( {x = - 3,y = 4} \right)\,{\rm{and}}\,\left( {x = 4,y = - 3} \right)$.
Note: In such types of questions, we can evaluate the values by changing the equation in a single variable. Always use BODMAS when there is a simplification problem.
Complete step-by-step answer:
Given equations are:
$5\left( {x + y} \right) + 2xy = - 19$
Apply BODMAS and solve the bracket first.
$\begin{array}{c}
5x + 5y + 2xy = - 19............{\rm{(1)}}\\
3xy + x + y = - 35............{\rm{(2)}}
\end{array}$
Multiply the equation (2) by $5$.
$\begin{array}{c}
5x + 5y + 2xy = - 19..............{\rm{(1)}}\\
15xy + 5x + 5y = - 175............{\rm{(3)}}
\end{array}$
Subtract equation (3) from equation (1).
$\begin{array}{c}
2xy - 15xy = - 19 - ( - 175)\\
- 13xy = 156\\
xy= - 12.................{\rm{(4)}}
\end{array}$
Substitute the value in equation (2) from equation (4)
$\begin{array}{c}
3\left( { - 12} \right) + x + y = - 35\\
x + y = 1\\
x = 1 - y...........(5)
\end{array}$
Substitute the value of $x$ in equation (4).
$\begin{array}{l}
\left( {1 - y} \right)y = - 12\\
{y^2} - y - 12 = 0
\end{array}$
$\begin{array}{l}
{y^2} - 4y + 3y - 12 = 0\\
y\left( {y - 4} \right) + 3\left( {y - 4} \right) = 0\\
\left( {y + 3} \right)\left( {y - 4} \right) = 0
\end{array}$
Evaluate further
$\begin{array}{l}
\left( {y + 3} \right) = 0,\left( {y - 4} \right) = 0\\
y = - 3,4..........{\rm{(6)}}
\end{array}$
Substitute the values of $y$ in equation (5).
$\begin{array}{l}
x = 1 - \left( { - 3} \right)\\
x = 4
\end{array}$
$\begin{array}{l}
x = 1 - 4\\
x = - 3
\end{array}$
So the answer is $\left( {x = - 3,y = 4} \right)\,{\rm{and}}\,\left( {x = 4,y = - 3} \right)$.
Note: In such types of questions, we can evaluate the values by changing the equation in a single variable. Always use BODMAS when there is a simplification problem.
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