Sure! To solve the pair of simultaneous equations:
[tex]\[
\begin{array}{c}
3x + y = 19 \quad \quad \text{(1)}\\
x - y = 1 \quad \quad \text{(2)}
\end{array}
\][/tex]
Here's a detailed step-by-step solution:
1. Isolate \( y \) in the second equation.
From equation (2):
[tex]\[
x - y = 1
\][/tex]
We can solve for \( y \) by adding \( y \) to both sides and then subtracting 1 from both sides:
[tex]\[
x - 1 = y
\][/tex]
So, we have:
[tex]\[
y = x - 1
\][/tex]
2. Substitute \( y = x - 1 \) into the first equation.
Replace \( y \) in equation (1) with \( x - 1 \):
[tex]\[
3x + (x - 1) = 19
\][/tex]
3. Simplify the resulting equation to find \( x \).
Combine like terms:
[tex]\[
3x + x - 1 = 19 \implies 4x - 1 = 19
\][/tex]
Next, add 1 to both sides:
[tex]\[
4x = 20
\][/tex]
Finally, divide both sides by 4:
[tex]\[
x = 5
\][/tex]
4. Substitute \( x = 5 \) back into \( y = x - 1 \).
Using the expression for \( y \):
[tex]\[
y = 5 - 1
\][/tex]
So, we get:
[tex]\[
y = 4
\][/tex]
5. Verify the solution.
Substitute \( x = 5 \) and \( y = 4 \) back into the original equations to ensure they are true:
For equation (1):
[tex]\[
3(5) + 4 = 15 + 4 = 19
\][/tex]
For equation (2):
[tex]\[
5 - 4 = 1
\][/tex]
Both original equations are satisfied with \( x = 5 \) and \( y = 4 \).
So, the solution to the system of equations is:
[tex]\[
(x, y) = (5, 4)
\][/tex]
