Substitution method
Now, let us take the following system of equations,
$\begin{align*} x+4y&=-1\\ 2x-3y&=9 \end{align*}$
and solve them by another method, called the substitution method, as follows.
Take one of the equations (I will take the first) and solve it for one variable. I take $x$, since it is easy to solve for:
$x=-1-4y$
Now, substitute this $x$ into the second equation; you will get
$2(-1-4y)-3y=9$
Solve this equation for $y$,
$-2-8y-3y=9$
Simplifying,
$-2-11y=9$
Add $+2$,
$-11y=11$
Divide by $-11$:
$y=-1$
Next, substitute this $y$ value into the equation we solved for $x$, $x=-1-4y$:
$x=-1-4\cdot (-1)$
Simplifying,
$x=3$
Now that you have the $x$ and $y$ values, write the solution as an ordered pair. The solution is $(3,-1)$.
Example 1: Solve the following system of equations by substitution.
$\begin{align*}5x+4y&=-10\\y&=2x-9\end{align*}$
Solution: The second equation is already solved for $y$, so substitute $2x-9$ for $y$ in the first equation:
$5x+4(2x-9)=-10$
Simplifying,
$\begin{align*}5x+8x-36&=-10\\13x-36&=-10\\13x&=26\\x&=2\end{align*}$
Substitute $x=2$ into $y=2x-9$:
$y=2(2)-9=-5$
The solution is $(2,-5)$.
If a system has no solution or infinitely many solutions, then in the second step the variable cancels, and you end up with a number on the left-hand side and another on the right-hand side. If the numbers are not the same — that is, the statement is false — then the system has no solution. If the numbers are the same — that is, the statement is true — then the system has infinitely many solutions.
Example 2: Solve the following system of equations
$\begin{align*} 5x+4 &=y\\ 5x-y &=-8\\ \end{align*}$
Solution:
In the first equation, $y$ is already isolated, so we do not have to solve for it. Just substitute $y$ into the second equation:
$5x-(5x+4)=-8$
Simplifying,
$-4=-8$
This statement is not true; therefore, the system of equations has no solution.
A system of equations with no solution is called an inconsistent system.
Example 3: Solve the following system of equations by substitution.
$\begin{align*}x&=-\dfrac{3}{7}y+4\\x+3y&=40\end{align*}$
Solution: The first equation is already solved for $x$, so substitute $-\dfrac{3}{7}y+4$ for $x$ in the second equation:
$-\dfrac{3}{7}y+4+3y=40$
Multiply every term by $7$ to clear the fraction:
$-3y+28+21y=280$
Simplifying,
$\begin{align*}18y+28&=280\\18y&=252\\y&=14\end{align*}$
Substitute $y=14$ into $x=-\dfrac{3}{7}y+4$:
$x=-\dfrac{3}{7}(14)+4=-6+4=-2$
The solution is $(-2,14)$.