After substitution you should get a quadratic in $y$. Use $b>2a$ to restrict whether that quadratic can have real solutions.

Once you’ve narrowed down the relationship between $a$ and $b$, plug $x=6$ into the original equation to finish.

Enter

• $y=\sqrt{2x-b}$
• $y=x-a$

and create sliders for $a$ and $b$.

Adjust $a$ and $b$ so that the graphs intersect at $x=6$ (the intersection point should have $x$-coordinate 6). Also ensure the square root is defined there, meaning $12-b\ge 0$.

While keeping an intersection at $x=6$, adjust so there is no second intersection. Use the graph window (or zoom out) to check that only one intersection appears. S⁠AT рrер‌ by‍ Anік‌o.aі

Once the settings show a single intersection at $x=6$ and satisfy $b>2a$, compute $a+b$ from your slider values and match it to the choices.

Let $y=\sqrt{2x-b}$. Then $y\ge 0$ and $2x-b=y^2$, so

Substitute into $y=x-a$:

Multiply by 2 and rearrange: © ​аnik‍o.‍aі

The quadratic in $y$ has discriminant

A real solution requires $\Delta\ge 0$, so $2a-b+1\ge 0$.

But the problem also gives $b>2a$, which implies $2a-b+1<1$.

Since $a$ and $b$ are integers, $2a-b+1$ is an integer. The only way it can be both $\ge 0$ and $<1$ is

Substitute $x=6$ into the original equation:

Now substitute $b=2a+1$:

Square both sides:

Bring all terms to one side:

So $a=5$, and then $b=2a+1=11$. Therefore,