Question

what is the solution to the linear equation?
d - 10 - 2d + 7 = 8 + d - 10 - 3d
d = -5
d = -1
d = 1
d = 5

Explanation:

Step1: Combine like terms on both sides

Left side: \(d - 2d - 10 + 7 = -d - 3\)
Right side: \(d - 3d + 8 - 10 = -2d - 2\)
So the equation becomes \(-d - 3 = -2d - 2\)

Step2: Add \(2d\) to both sides

\(-d + 2d - 3 = -2d + 2d - 2\)
\(d - 3 = -2\)

Step3: Add 3 to both sides

\(d - 3 + 3 = -2 + 3\)
\(d = 1\)

Wait, there's a mistake above. Let's redo Step1 correctly:

Step1: Combine like terms on both sides

Left side: \(d - 2d - 10 + 7 = (d - 2d) + (-10 + 7) = -d - 3\)
Right side: \(d - 3d + 8 - 10 = (d - 3d) + (8 - 10) = -2d - 2\)
Equation: \(-d - 3 = -2d - 2\)

Step2: Add \(2d\) to both sides

\(-d + 2d - 3 = -2d + 2d - 2\)
\(d - 3 = -2\)

Step3: Add 3 to both sides

\(d - 3 + 3 = -2 + 3\)
\(d = 1\)? Wait, no, let's check again. Wait, original equation:

Left side: \(d -10 -2d +7 = (d -2d) + (-10 +7) = -d -3\)

Right side: \(8 + d -10 -3d = (d -3d) + (8 -10) = -2d -2\)

So equation: \(-d -3 = -2d -2\)

Add \(2d\) to both sides: \(d -3 = -2\)

Add 3 to both sides: \(d = 1\)? But let's plug \(d = 1\) back into original equation:

Left: \(1 -10 -2(1) +7 = 1 -10 -2 +7 = (1 +7) + (-10 -2) = 8 -12 = -4\)

Right: \(8 +1 -10 -3(1) = 9 -10 -3 = -4\). So \(d = 1\) is correct. Wait, but let's check the options. One of the options is \(d = 1\). So the correct answer is \(d = 1\).

Wait, but when I first did Step3, I thought I made a mistake, but plugging in shows it's correct. So the correct solution is \(d = 1\).

Answer:

\(d = 1\) (Option: \(d = 1\))