The problem asks us to solve four systems of linear equations using the substitution method. System 1: $x + y = 4$ $2x + 3y = 11$ System 2: $2x - 5y = 1$ $3x + 2y = -8$ System 3: $3x - y = 3$ $2x + 5y = 19$ System 4: $3x - 5y = -4$ $7x - 8y = 9$

1. Problem Description

The problem asks us to solve four systems of linear equations using the substitution method.

System 1:

$x + y = 4$

$2x + 3y = 11$

System 2:

$2x - 5y = 1$

$3x + 2y = -8$

System 3:

$3x - y = 3$

$2x + 5y = 19$

System 4:

$3x - 5y = -4$

$7x - 8y = 9$

2. Solution Steps

System 1:

$x + y = 4$

$2x + 3y = 11$

From the first equation, we can express $x$ in terms of $y$:

$x = 4 - y$

Substitute this expression for $x$ into the second equation:

$2(4 - y) + 3y = 11$

$8 - 2y + 3y = 11$

$y = 3$

Now substitute $y = 3$ back into the equation $x = 4 - y$:

$x = 4 - 3 = 1$

So the solution is $x = 1$, $y = 3$.

System 2:

$2x - 5y = 1$

$3x + 2y = -8$

From the first equation, we can express $x$ in terms of $y$:

$2x = 1 + 5y$

$x = \frac{1 + 5y}{2}$

Substitute this expression for $x$ into the second equation:

$3(\frac{1 + 5y}{2}) + 2y = -8$

Multiply both sides by 2 to eliminate the fraction:

$3(1 + 5y) + 4y = -16$

$3 + 15y + 4y = -16$

$19y = -19$

$y = -1$

Now substitute $y = -1$ back into the equation $x = \frac{1 + 5y}{2}$:

$x = \frac{1 + 5(-1)}{2} = \frac{1 - 5}{2} = \frac{-4}{2} = -2$

So the solution is $x = -2$, $y = -1$.

System 3:

$3x - y = 3$

$2x + 5y = 19$

From the first equation, we can express $y$ in terms of $x$:

$y = 3x - 3$

Substitute this expression for $y$ into the second equation:

$2x + 5(3x - 3) = 19$

$2x + 15x - 15 = 19$

$17x = 34$

$x = 2$

Now substitute $x = 2$ back into the equation $y = 3x - 3$:

$y = 3(2) - 3 = 6 - 3 = 3$

So the solution is $x = 2$, $y = 3$.

System 4:

$3x - 5y = -4$

$7x - 8y = 9$

From the first equation, we can express $x$ in terms of $y$:

$3x = 5y - 4$

$x = \frac{5y - 4}{3}$

Substitute this expression for $x$ into the second equation:

$7(\frac{5y - 4}{3}) - 8y = 9$

Multiply both sides by 3 to eliminate the fraction:

$7(5y - 4) - 24y = 27$

$35y - 28 - 24y = 27$

$11y = 55$

$y = 5$

Now substitute $y = 5$ back into the equation $x = \frac{5y - 4}{3}$:

$x = \frac{5(5) - 4}{3} = \frac{25 - 4}{3} = \frac{21}{3} = 7$

So the solution is $x = 7$, $y = 5$.

3. Final Answer

System 1: $x = 1$, $y = 3$

System 2: $x = -2$, $y = -1$

System 3: $x = 2$, $y = 3$

System 4: $x = 7$, $y = 5$