Question 124·Easy·Nonlinear Equations in One Variable; Systems in Two Variables

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$x = -3$

$y = x^2 - 2x$

The graphs of the equations in the given system intersect at the point $(x, y)$ in the $xy$ -plane. What is the value of $y$ ?

In a system, the same point must satisfy both equations. If one equation already gives $x$ , substitute it into the equation for $y$ ; Desmos can evaluate that expression as a function. Keep parentheses around a negative $x$ : squaring it and multiplying it by a negative number both affect the signs.

Hints

Hint 1
A solution to a system makes both equations true. The first equation already fixes the $x$ -coordinate, so where should that value go in the equation for $y$ ?
Hint 2
Use substitution: replace every $x$ in the expression for $y$ with $(-3)$ . Keep the parentheses so you square the negative number and multiply $-2$ by the negative number.

Step-by-step

Substitute and evaluate
Step 1
Put the given x-value into the y-equation
A point where the graphs meet must satisfy both equations. Since the first gives $x=-3$ , replace each $x$ in the equation for $y$ with $-3$ :
$y=(-3)^2-2(-3).$
This is substitution. Keep the parentheses: $-2$ times a negative number is positive.
Step 2
Evaluate to find y
In Desmos, type $f(x)=x^2-2x$ , then $f(-3)$ . Desmos prints $15$ . Here $f(-3)$ evaluates the expression for $y$ at the given $x$ , so the $y$ -coordinate is $15$ . Choice C.

Question 124·Easy·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