Here's the question you clicked on:
2yoututu11
Solve the following system of equations by linear combination: 2d + e = 8 d – e = 4 A. The solution is (5, –2). B. The solution is (4, 0). C. There is no solution. D. There are an infinite number of solutions.
Answer is B 2D+E=8 D-E=4 D=4+E substitute it into first equation 2(4+E)+E=8 expand brackets 8+2E+E=8 3E=8-8 3E=0 E=0 substitute into original equation D-0=4 D=4 (4,0)