Answer :
Answer:
Let's denote the two unknown numbers as \( x \) and \( y \).
According to the problem, we have two equations:
1. Two times the first number and three times the second number makes 120:
\[
2x + 3y = 120
\]
2. Three times the second number subtracted from four times the first number gives 78:
\[
4x - 3y = 78
\]
To find the values of \( x \) and \( y \), we will solve these two simultaneous linear equations.
First, we add the two equations to eliminate \( y \):
\[
2x + 3y = 120 \quad \text{(Equation 1)}
\]
\[
4x - 3y = 78 \quad \text{(Equation 2)}
\]
Adding these equations:
\[
(2x + 3y) + (4x - 3y) = 120 + 78
\]
\[
2x + 4x + 3y - 3y = 198
\]
\[
6x = 198
\]
\[
x = \frac{198}{6}
\]
\[
x = 33
\]
Now, substitute \( x = 33 \) back into one of the original equations to solve for \( y \). We use Equation 1:
\[
2x + 3y = 120
\]
\[
2(33) + 3y = 120
\]
\[
66 + 3y = 120
\]
\[
3y = 120 - 66
\]
\[
3y = 54
\]
\[
y = \frac{54}{3}
\]
\[
y = 18
\]
Thus, the two numbers are:
\[
x = 33
\]
\[
y = 18
\]
### Verification:
- For the first equation: \( 2x + 3y = 2(33) + 3(18) = 66 + 54 = 120 \), which is correct.
- For the second equation: \( 4x - 3y = 4(33) - 3(18) = 132 - 54 = 78 \), which is also correct.
So, the numbers are:
\[
\boxed{33 \text{ and } 18}
\]