To solve the given system of linear equations:
[tex]\[
\begin{cases}
2x + 3y = 30 \\
4x + y = 50
\end{cases}
\][/tex]
we will use the method of substitution or elimination. Here, I'll use the method of elimination for a detailed step-by-step solution.
### Step 1: Align the Equations
First, we note the two equations:
1. [tex]\( 2x + 3y = 30 \)[/tex]
2. [tex]\( 4x + y = 50 \)[/tex]
### Step 2: Eliminate One Variable
To eliminate one variable, we can manipulate the equations so that either the coefficients of [tex]\(x\)[/tex] or [tex]\(y\)[/tex] become the same (or multiples thereof). Here, we shall eliminate [tex]\(y\)[/tex].
To do this, we can multiply the second equation by 3 to align the coefficients of [tex]\(y\)[/tex] with those in the first equation.
[tex]\[
4x + y = 50 \quad \text{(original second equation)}
\][/tex]
Multiply through by 3:
[tex]\[
3 \cdot (4x + y) = 3 \cdot 50
\][/tex]
This gives:
[tex]\[
12x + 3y = 150
\][/tex]
Now we have the system:
1. [tex]\( 2x + 3y = 30 \)[/tex]
2. [tex]\( 12x + 3y = 150 \)[/tex]
### Step 3: Subtract the Equations
To eliminate [tex]\(y\)[/tex], subtract the first equation from the newly obtained second equation:
[tex]\[
(12x + 3y) - (2x + 3y) = 150 - 30
\][/tex]
This simplifies to:
[tex]\[
10x = 120
\][/tex]
### Step 4: Solve for [tex]\(x\)[/tex]
Divide both sides by 10:
[tex]\[
x = \frac{120}{10} = 12
\][/tex]
So, [tex]\(x = 12\)[/tex].
### Step 5: Substitute [tex]\(x\)[/tex] Back into One of the Original Equations
Now that we have [tex]\(x = 12\)[/tex], we substitute this value back into the second original equation to find [tex]\(y\)[/tex].
Using [tex]\( 4x + y = 50 \)[/tex]:
[tex]\[
4(12) + y = 50
\][/tex]
This simplifies to:
[tex]\[
48 + y = 50
\][/tex]
Subtract 48 from both sides:
[tex]\[
y = 50 - 48 = 2
\][/tex]
So, [tex]\(y = 2\)[/tex].
### Step 6: Verify the Solution
Finally, we verify our solution by substituting [tex]\(x = 12\)[/tex] and [tex]\(y = 2\)[/tex] back into the original equations.
1. [tex]\( 2x + 3y = 30 \)[/tex]:
[tex]\[
2(12) + 3(2) = 24 + 6 = 30 \quad \text{(checks out)}
\][/tex]
2. [tex]\( 4x + y = 50 \)[/tex]:
[tex]\[
4(12) + 2 = 48 + 2 = 50 \quad \text{(checks out)}
\][/tex]
Thus, the solution to the system of equations is:
[tex]\[
(x, y) = (12, 2)
\][/tex]
