Solve the system of equations: What is the solution ?

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SAT Nonlinear Equations & Systems Question 148 | Free SAT Prep | Aniko

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

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Solve the system of equations:

\begin{aligned} y &= |x-2| + 1 \\ y &= -x + 7 \end{aligned}

What is the solution $(x,y)$ ?

For systems involving an absolute value and a line, the most reliable algebraic method is to set the two expressions equal and break the absolute value into cases (e.g., $x$ above or below a key point like $2$ ). Solve each case, discard any $x$ -values that violate the case condition, then plug valid $x$ -values back into an easier equation (usually the line) to find $y$ . On multiple-choice questions, you can also check each answer choice quickly by substituting the $(x,y)$ pair into both equations and seeing which one satisfies both.

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Connect the two equations

Since both equations equal $y$ , try setting $|x-2| + 1$ equal to $-x + 7$ and solve that equation.

Handle the absolute value

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

Check your solutions and use them to find y

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.

Desmos Guide

Enter the two equations

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.

Locate the intersection point

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.

Match the coordinates to an answer choice

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

Step-by-step Explanation

Set the equations equal

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

|x-2| + 1 = -x + 7.

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

Use cases for the absolute value

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

x - 2 + 1 = -x + 7 \\[0.7em] x - 1 = -x + 7.

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

2x = 8 \Rightarrow x = 4.

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

Case 2: $x < 2$

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

-x + 2 + 1 = -x + 7 \\[0.7em] -x + 3 = -x + 7.

Add $x$ to both sides:

3 = 7,

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

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

Find the corresponding y-value and state the solution

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

y = -4 + 7 = 3.

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

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Step-by-step Explanation

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Desmos Guide

Step-by-step Explanation