The system of equations below is given. What is one possible value of ?

Solution available with step-by-step explanation

SAT Nonlinear Equations & Systems Question 220 | Free SAT Prep | Aniko

00:00

Question 220·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

0 / 233

The system of equations below is given.

\begin{aligned} x^2 + y^2 &= 25 \\ y &= x + 1 \end{aligned}

What is one possible value of $x$ ?

For systems where a line intersects a circle (or another nonlinear graph), first use substitution: replace one variable using the linear equation to get a single equation in the other variable. Simplify this to a standard quadratic form, then solve it quickly by factoring if possible (or use the quadratic formula when factoring is hard). Always check for multiple solutions and remember that if the question asks for “one possible value,” any value that satisfies both original equations is acceptable.

Hints

Click me!

Combine the equations

You know that $y = x + 1$ . How can you use this equation to replace $y$ in the other equation?

Form a single-variable equation

After you substitute for $y$ in $x^2 + y^2 = 25$ , you should get an equation involving only $x$ . What type of equation is it?

Solve the quadratic

Rewrite your equation in the form $ax^2 + bx + c = 0$ , then solve it by factoring. Make sure you find both values of $x$ that work.

Desmos Guide

Graph the circle

In Desmos, enter the equation x^2 + y^2 = 25. This will draw the circle of radius $5$ centered at the origin.

Graph the line

On a new line, enter y = x + 1. This will draw the line that must intersect the circle at one or more points.

Find the intersection points

Use the intersection tool (or click where the line crosses the circle). Desmos will show the coordinates of the intersection points; note the $x$ -coordinates of these points—each is a possible value of $x$ that solves the system.

Step-by-step Explanation

Use substitution to get one equation in one variable

We are given the system

\begin{aligned} x^2 + y^2 &= 25 \\ y &= x + 1 \end{aligned}

Since $y=x+1$ , substitute $x+1$ for $y$ in the first equation:

x^2 + (x+1)^2 = 25.

Expand and simplify the equation

Now expand $(x+1)^2$ and combine like terms:

\begin{aligned} x^2 + (x+1)^2 &= 25 \\ x^2 + (x^2 + 2x + 1) &= 25 \\ 2x^2 + 2x + 1 &= 25. \end{aligned}

Subtract $25$ from both sides to get a standard quadratic form:

\begin{aligned} 2x^2 + 2x + 1 - 25 &= 0 \\ 2x^2 + 2x - 24 &= 0. \end{aligned}

Finally, divide everything by $2$ to simplify:

x^2 + x - 12 = 0.

Factor the quadratic

Now factor the quadratic $x^2 + x - 12$ .

We look for two numbers that multiply to $-12$ and add to $1$ . Those numbers are $4$ and $-3$ .

So the factorization is

x^2 + x - 12 = (x + 4)(x - 3) = 0.

Solve for x and choose one possible value

From $(x + 4)(x - 3) = 0$ , set each factor equal to zero:

\begin{aligned} x + 4 &= 0 \\ x - 3 &= 0 \end{aligned}

\begin{aligned} x &= -4 \\ x &= 3. \end{aligned}

Both values satisfy the system, and the question asks for one possible value of $x$ . A correct response is $3$ .

Strategy

Hints

Desmos Guide

Step-by-step Explanation

Study Hints

Strategy

Hints

Desmos Guide

Step-by-step Explanation