To solve the system of equations:
- \( 2x + 3y = 38 \) (Equation 1)
- \( 3x + 22y = 37 \) (Equation 2)
We can use the substitution or elimination method. Here, I'll use the elimination method.
First, we will express one variable in terms of the other using Equation 1.
From Equation 1: \[ 2x + 3y = 38 \] We can isolate \( x \): \[ 2x = 38 - 3y \] \[ x = 19 - \frac{3y}{2} \]
Next, we substitute this expression for \( x \) into Equation 2.
Substituting into Equation 2: \[ 3\left(19 - \frac{3y}{2}\right) + 22y = 37 \] Distributing \( 3 \): \[ 57 - \frac{9y}{2} + 22y = 37 \] Now, we can get rid of the fraction by multiplying the entire equation by \( 2 \): \[ 2 \cdot 57 - 9y + 44y = 2 \cdot 37 \] This simplifies to: \[ 114 - 9y + 44y = 74 \] Combining like terms gives: \[ 114 + 35y = 74 \] Next, subtract \( 114 \) from both sides: \[ 35y = 74 - 114 \] \[ 35y = -40 \] Dividing by \( 35 \): \[ y = -\frac{40}{35} = -\frac{8}{7} \]
Now, we will substitute \( y \) back into the expression for \( x \): \[ x = 19 - \frac{3(-\frac{8}{7})}{2} \] This simplifies as follows: \[ x = 19 + \frac{24}{14} \] Converting \( \frac{24}{14} \) to its simplest form gives: \[ x = 19 + \frac{12}{7} = \frac{133}{7} + \frac{12}{7} = \frac{145}{7} \]
Thus, the solution to the system of equations is: \[ x = \frac{145}{7}, \quad y = -\frac{8}{7} \]
In decimal form, this can be approximated as: \[ x \approx 20.71, \quad y \approx -1.14 \]