Question

question 11
solve the equation for the given variable.
$2x + 7 - 2x + 8 = 0x + 15$
answer: $x = $

Explanation:

Step1: Combine like terms on left

\(2x - 2x + 7 + 8 = 0x + 15\)
Simplifies to \(0x + 15 = 0x + 15\)

Step2: Analyze the equation

The equation \(0x + 15 = 0x + 15\) means \(15 = 15\), which is always true. So \(x\) can be any real number. But since the original equation's \(x\) terms cancel, and constants match, it's an identity (all real numbers are solutions). However, if we follow the process:
After combining, we have \(15 = 15\), so no restriction on \(x\), but in the context of solving, since the \(x\) terms eliminate and the equation is always true, \(x\) is all real numbers. But maybe the problem expects recognizing it's an identity (infinite solutions) or maybe a typo, but as is, let's see:
Wait, let's re - do:
Left side: \(2x - 2x + 7 + 8=(2x - 2x)+(7 + 8)=0 + 15 = 15\)
Right side: \(0x+15 = 15\)
So \(15 = 15\), which is true for all \(x\in\mathbb{R}\). But if we consider the problem's format, maybe it's a trick question. But if we have to write, since the equation is always true, \(x\) can be any real number. But maybe the problem expects "all real numbers" or maybe there's a mistake. But let's check again.

Wait, maybe the problem is written wrong? But as per the given equation:

\(2x+7 - 2x + 8=0x + 15\)

Simplify left: \( (2x-2x)+(7 + 8)=15\)

Right: \(0x + 15=15\)

So \(15 = 15\), which is an identity, so \(x\) is any real number. But if we have to put in the box, maybe the problem expects "all real numbers" or maybe it's a mistake. But perhaps the original problem had a typo, but as is, the solution is all real numbers. However, if we consider that maybe the equation is supposed to have a unique solution, but in this case, it's an identity.

But let's see, maybe the problem is intended to have a unique solution, but due to the way it's written, the \(x\) terms cancel. So the answer is that \(x\) can be any real number. But if we have to write in the box, maybe the problem expects "all real numbers" or maybe there's an error. But according to the equation, it's an identity.

Answer:

All real numbers (or \(x\in\mathbb{R}\))