The system of equations above relates the variables , , and , where and . Which of the following expresses in terms of ?

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SAT Nonlinear Equations & Systems Question 63 | Free SAT Prep | Aniko

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Question 63·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

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\begin{cases} xy = 6 \\ 2x + 3y = k \end{cases}

The system of equations above relates the variables $x$ , $y$ , and $k$ , where $x \neq 0$ and $y \neq 0$ . Which of the following expresses $x$ in terms of $k$ ?

When a system mixes a product like $xy=\text{constant}$ with a linear combination like $ax+by=k$ , first solve the product equation for one variable and substitute into the linear equation. Clear any fractions by multiplying through, then rewrite the resulting equation as a quadratic in the remaining variable and apply the quadratic formula carefully, paying close attention to the discriminant and keeping the $\pm$ to represent both possible solutions.

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Start with substitution

Use $xy = 6$ to solve for one variable (either $x$ in terms of $y$ or $y$ in terms of $x$ ), then substitute into the other equation.

Eliminate the fraction

After substituting into $2x + 3y = k$ , you will get a fraction. Multiply both sides of the equation by $x$ to clear the denominator.

Recognize the quadratic in x

Once you clear the fraction, rearrange the equation into the form $ax^2 + bx + c = 0$ and use the quadratic formula. Be careful computing the discriminant $b^2 - 4ac$ .

Remember both solutions

When you apply the quadratic formula, keep the $\pm$ ; this represents the two possible $x$ -values that can solve the system.

Desmos Guide

Graph the system for a specific k-value

Pick a convenient value for $k$ that makes $k^2 - 144$ positive (for example, $k = 20$ ). In Desmos, enter the two equations: y = 6/x and 2x + 3y = 20. Look at the intersection points and note their $x$ -coordinates.

Turn each answer choice into an x-expression

In Desmos, define functions for the $x$ -values given by each option, for example A(k) = (k + sqrt(k^2 - 144))/4, B(k) = (k - sqrt(k^2 - 144))/4, etc. Then evaluate each at $k = 20$ to get numerical $x$ -values.

Match the intersections to the answer choice

Compare the $x$ -coordinates of the intersection points from the graph of the system with the values you get from each answer choice at $k = 20$ . The correct choice is the one whose expression produces exactly the $x$ -values that solve the system.

Optionally test another k

If you want extra confirmation, choose a different $k$ that still makes the square root real, graph the system with that new $k$ , and repeat the comparison. The correct expression will match the system’s solutions for every valid $k$ you try.

Step-by-step Explanation

Express one variable in terms of the other

From the first equation, $xy = 6$ , and the condition $x \neq 0$ , we can solve for $y$ :

y = \frac{6}{x}

Now substitute this expression for $y$ into the second equation $2x + 3y = k$ .

Substitute and form a quadratic equation in x

Substitute $y = 6/x$ into $2x + 3y = k$ :

2x + 3\left(\frac{6}{x}\right) = k

This simplifies to:

2x + \frac{18}{x} = k

Multiply every term by $x$ (allowed because $x \neq 0$ ):

2x^2 + 18 = kx

Rearrange all terms to one side to get a standard quadratic in $x$ :

2x^2 - kx + 18 = 0

Use the quadratic formula to solve for x

For the quadratic equation

2x^2 - kx + 18 = 0,

the coefficients are $a = 2$ , $b = -k$ , and $c = 18$ .

The quadratic formula is

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

Compute the discriminant:

\begin{aligned} b^2 - 4ac &= (-k)^2 - 4(2)(18) \\ &= k^2 - 144. \end{aligned}

Now plug into the quadratic formula:

\begin{aligned} x &= \frac{-(-k) \pm \sqrt{k^2 - 144}}{2 \cdot 2} \\ &= \frac{k \pm \sqrt{k^2 - 144}}{4}. \end{aligned}

So $x$ in terms of $k$ is

\frac{k \pm \sqrt{k^2 - 144}}{4}.

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Step-by-step Explanation

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