Solve the system of equations using the linear combination method.

[tex]
\begin{cases}
2d - e = 7 \\
d + e = 5
\end{cases}
[/tex]

A. There is no solution.
B. The solution is [tex]$(4,1)$[/tex].
C. There are infinitely many solutions.
D. The solution is [tex]$(3,-1)$[/tex].



Answer :

To solve the system of linear equations using the linear combination method (or elimination method), we proceed as follows:

### Given System of Equations:
[tex]\[ \begin{cases} 2d - e = 7 \\ d + e = 5 \end{cases} \][/tex]

### Step-by-Step Solution:

1. Write down the equations:
[tex]\[ \text{Equation (1): } 2d - e = 7 \][/tex]
[tex]\[ \text{Equation (2): } d + e = 5 \][/tex]

2. Choose one variable to eliminate:
We'll eliminate [tex]\(e\)[/tex].

3. Add the two equations together to eliminate [tex]\(e\)[/tex]:

Let's add Equation (1) and Equation (2):
[tex]\[ (2d - e) + (d + e) = 7 + 5 \][/tex]

4. Simplify the resulting equation:
[tex]\[ 2d - e + d + e = 12 \][/tex]
[tex]\[ 3d = 12 \][/tex]

5. Solve for [tex]\(d\)[/tex]:
[tex]\[ d = \frac{12}{3} = 4 \][/tex]

6. Substitute the value of [tex]\(d\)[/tex] back into one of the original equations:
We use Equation (2) for this:
[tex]\[ 4 + e = 5 \][/tex]

7. Solve for [tex]\(e\)[/tex]:
[tex]\[ e = 5 - 4 = 1 \][/tex]

So, we find:
[tex]\[ d = 4, \quad e = 1 \][/tex]

### Conclusion:
The solution to the system of equations is [tex]\((d, e) = (4,1)\)[/tex]. This can be written as:
[tex]\[ \boxed{(4,1)} \][/tex]