Answer
True statement
Work Step by Step
Suppose a first-degree equation is
$y=2x+1$ …… (1)
And a second-degree equation is
${{y}^{2}}=xy+3$ …… (2)
Now, solve using equations 1 and 2:
$\begin{align}
& {{y}^{2}}=xy+3 \\
& {{\left( 2x+1 \right)}^{2}}=x\left( 2x+1 \right)+3 \\
& 4{{x}^{2}}+1+4x=2{{x}^{2}}+x+3 \\
& \left( 4{{x}^{2}}-2{{x}^{2}} \right)+\left( 4x-x \right)+\left( 1-3 \right)=0
\end{align}$
Simplify:
$2{{x}^{2}}+3x-2=0$
Factor the equation,
$\begin{align}
& \left( 2x-1 \right)\left( x+2 \right)=0 \\
& \\
\end{align}$
Simplify both the factors,
$\begin{align}
& 2x-1=0 \\
& 2x=1 \\
& x=\frac{1}{2}
\end{align}$
And
$\begin{align}
& x+2=0 \\
& x=-2
\end{align}$
Substitute $x=\frac{1}{2}$ in equation (2), and solve for y:
$\begin{align}
& y=2x+1 \\
& =2\left( \frac{1}{2} \right)+1 \\
& =1+1 \\
& =2
\end{align}$
The ordered pair is $\left( \frac{1}{2},2 \right)$.
Substitute $x=-2$ in equation (1), and solve for y:
$\begin{align}
& {{y}^{2}}=xy+3 \\
& {{y}^{2}}=\left( -2 \right)y+3 \\
& {{y}^{2}}+2y-3=0 \\
& \left( y-1 \right)\left( y+3 \right)=0
\end{align}$
Simplify both the factors,
$\begin{align}
& y-1=0 \\
& y=1
\end{align}$
And
$\begin{align}
& y+3=0 \\
& y=-3
\end{align}$
The ordered pairs are $\left( -2,1 \right)$ and $\left( -2,-3 \right)$.
Thus, the systems containing one first-degree equation and one second-degree equation are most easily solved using the substitution method.
