A system of equations is given by If is a solution to the system, what is one possible value of ?

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SAT Nonlinear Equations & Systems Question 92 | Free SAT Prep | Aniko

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Question 92·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

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A system of equations is given by

\begin{aligned} x + y &= 10 \\ x^2 + y &= 34 \end{aligned}

If $(x,y)$ is a solution to the system, what is one possible value of $y$ ?

For systems where one equation is linear and the other is nonlinear (such as involving $x^2$ ), the fastest and most reliable method is substitution: solve the linear equation for one variable, substitute that expression into the nonlinear equation to get a single-variable equation, and then solve it carefully (using the quadratic formula if needed). Finally, substitute each solution back into the linear equation to find the corresponding values of the other variable, and remember that any resulting $y$ -value that satisfies both original equations is an acceptable answer when the problem asks for 'one possible value.'

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Use the simpler (linear) equation first

Focus on the equation $x + y = 10$ . Can you rewrite it to express $y$ in terms of $x$ ?

Substitute into the second equation

Once you have $y$ written in terms of $x$ , replace $y$ in the equation $x^2 + y = 34$ with that expression. What kind of equation in $x$ do you get?

Solve the equation in one variable

After substitution, you should have a quadratic equation in $x$ . If it does not factor nicely, what formula can you use to find the values of $x$ ?

Go back to find y

When you find each value of $x$ , plug it back into $y = 10 - x$ to find the corresponding $y$ values. Any of these $y$ values can answer the question.

Desmos Guide

Enter both equations in Desmos

Type the equations in function form:

y = 10 - x
y = 34 - x^2

Find the intersection points

Look for the points where the two graphs intersect. You can tap or click on the intersection points in Desmos to see their coordinates $(x, y)$ .

Read off the y-values

The $y$ -coordinates of the intersection points are the possible values of $y$ that satisfy the system. Either of these $y$ -values answers the question.

Step-by-step Explanation

Solve the linear equation for one variable

Start with the first equation:

x + y = 10

Solve for $y$ by subtracting $x$ from both sides:

y = 10 - x.

Now you have $y$ expressed in terms of $x$ .

Substitute into the second equation

Use $y = 10 - x$ in the second equation $x^2 + y = 34$ .

Substitute:

x^2 + (10 - x) = 34.

Simplify the left-hand side:

x^2 - x + 10 = 34.

Subtract 34 from both sides to set the equation equal to 0:

x^2 - x - 24 = 0.

Now you have a quadratic equation in $x$ .

Solve the quadratic equation for x

The quadratic equation is

x^2 - x - 24 = 0.

It does not factor nicely with integers, so use the quadratic formula with $a = 1$ , $b = -1$ , and $c = -24$ :

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.

Substitute the values of $a$ , $b$ , and $c$ :

x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-24)}}{2(1)} \\[0.7em] = \frac{1 \pm \sqrt{1 + 96}}{2} \\[0.7em] = \frac{1 \pm \sqrt{97}}{2}.

So there are two possible $x$ values:

x = \frac{1 + \sqrt{97}}{2} \quad \text{or} \quad x = \frac{1 - \sqrt{97}}{2}.

Find the corresponding y-values

Recall that $y = 10 - x$ .

For $x = \dfrac{1 + \sqrt{97}}{2}$ :

y = 10 - \frac{1 + \sqrt{97}}{2} = \frac{20 - 1 - \sqrt{97}}{2} = \frac{19 - \sqrt{97}}{2}.

For $x = \dfrac{1 - \sqrt{97}}{2}$ :

y = 10 - \frac{1 - \sqrt{97}}{2} = \frac{20 - 1 + \sqrt{97}}{2} = \frac{19 + \sqrt{97}}{2}.

Thus the two possible values of $y$ that satisfy the system are

y = \frac{19 - \sqrt{97}}{2} \quad \text{and} \quad y = \frac{19 + \sqrt{97}}{2}.

Since the problem asks for one possible value of $y$ , a correct response is, for example,

y = \frac{19 - \sqrt{97}}{2}.

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation