Solving systems of equations by substitution: 3 worked examples

We are given a system of two linear equations:

1. $y = 2x + 1$
2. $3x + y = 16$

Since the first equation is already solved for $y$, we can substitute $2x + 1$ for $y$ in the second equation:
$3x + (2x + 1) = 16$

Now, combine the like terms involving $x$:
$5x + 1 = 16$

Subtract $1$ from both sides of the equation:
$5x = 15$

Divide both sides by $5$:
$x = 3$

Now that we have the value of $x$, substitute it back into the first equation to solve for $y$:
$y = 2(3) + 1$
$y = 6 + 1$
$y = 7$

The solution to the system is the ordered pair $(3, 7)$.

We are given the system of linear equations:

1. $x = y - 4$
2. $2x + 3y = 22$

Since the first equation already expresses $x$ in terms of $y$, we can use the substitution method. Substitute $x = y - 4$ into the second equation:
$2(y - 4) + 3y = 22$

Distribute the $2$:
$2y - 8 + 3y = 22$

Combine the $y$ terms:
$5y - 8 = 22$

Add $8$ to both sides of the equation:
$5y = 30$

Divide by $5$:
$y = 6$

Now, substitute the value of $y$ back into the first equation to solve for $x$:
$x = 6 - 4$
$x = 2$

The solution to the system is $x = 2$ and $y = 6$. We can verify this by plugging the values into the second equation: $2(2) + 3(6) = 4 + 18 = 22$, which is correct.

We are given the system of equations:

1. $y = -x + 8$
2. $4x - y = 7$

Since the first equation is already solved for $y$, we can substitute $y = -x + 8$ into the second equation:
$4x - (-x + 8) = 7$

Distribute the negative sign:
$4x + x - 8 = 7$

Combine like terms:
$5x - 8 = 7$

Add $8$ to both sides of the equation:
$5x = 15$

Divide by $5$:
$x = 3$

Now substitute $x = 3$ back into the first equation to find $y$:
$y = -(3) + 8$
$y = 5$

The solution to the system is $(x, y) = (3, 5)$.