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 $k$ ?

The product $xy$ signals a nonlinear system. To express $x$ in terms of $k$ , replace $y$ using $xy=6$ , then solve the resulting quadratic exactly. Desmos can check whether a particular $k$ gives two solutions, but a graph alone can't produce a formula in $k$ . Watch for the two possible roots: neither can be dropped without a condition that rules it out.

Hints

Hint 1
Since $x$ is nonzero, divide $xy=6$ by $x$ to write $y$ in terms of $x$ . What happens when you replace $y$ in the other equation with that expression?
Hint 2
Multiply every term by $x$ to clear the fraction. Then put the equation in standard form, $ax^2+bx+c=0$ , so you can identify $a$ , $b$ , and $c$ with their signs.
Hint 3
The quadratic formula uses $\pm$ because adding and subtracting the square root can give two values. Does either nonzero condition in the question tell you to keep only one?

Step-by-step

Substitute, then use the quadratic formula
Step 1
Write y in terms of x
The given $xy=6$ and nonzero $x$ let you divide by $x$ :
$y=\frac{6}{x}.$
Step 2
Replace y in the other equation
Use substitution: put the whole fraction $\frac{6}{x}$ wherever the second equation has $y$ :
$2x+3\left(\frac{6}{x}\right)=k.$
Multiply $3$ by $6$ :
$2x+\frac{18}{x}=k.$
Step 3
Clear the denominator
Multiply every term by $x$ to remove the fraction. This preserves the equation because the problem says $x\ne0$ :
$2x^2+18=kx.$
Step 4
Put the quadratic in standard form
Subtract $kx$ from both sides:
$2x^2-kx+18=0.$
In standard form $ax^2+bx+c=0$ , the coefficients are $a=2$ , $b=-k$ , and $c=18$ . Keep the minus sign on $b$ .
Step 5
Calculate the number subtracted under the square root
The discriminant is $b^2-4ac$ , the part under the quadratic formula's square root. Type $4(2)(18)$ in Desmos. It prints $144$ , which must be subtracted from $b^2$ .
Step 6
Check whether both roots can count
For a two-root check, pick $(2,3)$ because $2\cdot3=6$ . Type $k=2(2)+3(3)$ , then graph both given equations in Desmos. It sets $k=13$ ; click the intersections to see $(2,3)$ and about $(4.5,1.333)$ . Both satisfy the nonzero conditions, so keeping only one $x$ would miss a solution.
Step 7
Keep both exact values of x
For $ax^2+bx+c=0$ , the quadratic formula is $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ . The $\pm$ means add and subtract the square root. Substitute the coefficients and the $144$ you found:
$x=\frac{-(-k)\pm\sqrt{(-k)^2-144}}{2(2)}.$
Simplify the signs and the square:
$x=\frac{k\pm\sqrt{k^2-144}}{2(2)}.$
Multiply the denominator:
$x=\frac{k\pm\sqrt{k^2-144}}{4}.$
These are the possible values of $x$ in terms of $k$ whenever the system has real solutions. Keep both roots unless a given condition rules one out. Choice C.

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

Strategy

Hints

Step-by-step

Lessons that teach this

Study Hints

Strategy

Hints

Step-by-step

Lessons that teach this