Question 148·Easy·Nonlinear Equations in One Variable; Systems in Two Variables

Remember that $|x-2|$ becomes $x-2$ when $x \ge 2$ and $-(x-2)$ when $x < 2$. Solve the equation separately for each case.

After you find any possible $x$-values, make sure they fit the case you assumed, then plug them back into one of the original equations to get $y$ and match the ordered pair to an answer choice.

Type y = abs(x - 2) + 1 for the absolute value graph and y = -x + 7 for the line. Make sure abs has parentheses around x - 2.

Look at the graph for the point where the V-shaped absolute value graph and the downward-sloping line cross. Click on the intersection; Desmos will display its coordinates.

Compare the intersection coordinates shown by Desmos to the listed answer choices and select the ordered pair that exactly matches those coordinates.

Both equations equal $y$, so set the right-hand sides equal to each other:

Now we need to solve this equation for $x$.

The expression $|x-2|$ depends on whether $x$ is at least $2$ or less than $2$.

Case 1: $x \ge 2$

Then $|x-2| = x-2$, so

Add $x$ to both sides and add $1$ to both sides:

This value satisfies $x \ge 2$, so it is valid.

Case 2: $x < 2$

Then $|x-2| = -(x-2) = -x + 2$, so

Add $x$ to both sides:

which is impossible. So there is no solution from this case.

Therefore, the only possible $x$-value from the system is $x = 4$.

Substitute $x = 4$ into either original equation to find $y$. Using the linear equation $y = -x + 7$:

So the solution to the system is the point $(4,3)$, which matches answer choice D.