To solve the system of equations given by \(4x + 3y = 15\) and \(y = x + 5\), we can use substitution or elimination. Let's use substitution, since the second equation is already solved for \(y\).
Step 1: Substitute \(y\) in the first equation
We know from the second equation: \[ y = x + 5 \]
Now substitute \(y\) in the first equation: \[ 4x + 3(x + 5) = 15 \]
Step 2: Simplify and solve for \(x\)
Distribute the 3: \[ 4x + 3x + 15 = 15 \] Combine like terms: \[ 7x + 15 = 15 \] Now, subtract 15 from both sides: \[ 7x = 0 \] Divide by 7: \[ x = 0 \]
Step 3: Substitute \(x\) back to find \(y\)
Using \(y = x + 5\): \[ y = 0 + 5 = 5 \]
Solution
The intersection point (solution) of the two equations is: \[ \boxed{(0, 5)} \]