Solve linear and nonlinear systems

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Difficulty: Advanced
Estimated time: 29 minutes
Domains: Advanced Math
Techniques: Nonlinear-systemsSubstitutionIntersectionsSolution-countTangency

What you’ll learn

Spot a system where at least one equation isn't a line, like a parabola or a circle.
Solve it by substitution, or by setting two expressions equal, when one variable is already by itself.
Read the answer from the graph's intersections when a decimal answer is fine.
Find the constant that makes two graphs meet zero, one, or two times.
Give exactly what the question asks for: a coordinate, a point, a value, or the number of solutions.

Prerequisites

Why this matters on the SAT

Find every point shared by a line and a curve

A solution to a system is a point $(x,y)$ that makes both equations true. On a graph, that's a spot where the two graphs meet. So when one equation is a line and the other is a curve, each intersection is one solution.

SAT example

The graphs of

y=x^2-2

and

y=3x-4

intersect at two points. What is the greater possible value of $y$ ?

A
$-1$
B
$1$
C
$3$
D
$2$

Solution to the example

Type each equation on its own line and click both intersections. They're

(1,-1)\qquad\text{and}\qquad(2,2).

The possible $y$ -coordinates are $-1$ and $2$ . The greater one is $\boxed{2}$ , so the answer is D.

The graph gave you both points. But the question's last sentence asks only for the greater $y$ -coordinate, the second number in each pair, so that's all you submit.

Common mistake:

It’s easy to click the right point and then type its x-coordinate. Once you have each point, label its two numbers x and y, then reread what the question asks for.

Calculator loads as you approach

Click both intersections. Each one is a point that solves both equations.

Recognize the system and choose a method

You're looking at this kind of system when two equations use the same two variables and at least one of them isn't a line. The giveaway is a squared variable, a square root, an absolute value, two variables multiplied together, or a circle or other curve.

A line and a parabola can meet zero, one, or two times, and two curves can sometimes meet even more often. So don't expect one tidy solution. Plan on finding them all.

Before you calculate, look at two things: how the equations are written, and what the question wants back.

Substitute by hand when…

One equation already has a variable by itself, like $y=x+3$ , so you can drop that expression into the other equation.
Both equations start with $y=$ (or both with $x=$ ), so you can set the right sides equal to each other.
The quadratic you get factors cleanly, like $x^2-3x-4=(x-4)(x+1)$ .
The answer has to be exact, like a fraction or a square root.

Start with the graph when…

The equations are easy to type in, and a decimal answer is fine, as with $y=0.4x+2.2$ .
The graphs might meet several times, and you want to see every meeting point at once.
The question only asks how many solutions there are, so counting intersections answers it.
The answer choices are exact, but each one works out to a different decimal you can match to a graph point.

When a question asks for the constant that gives a certain number of solutions, like exactly one, use a hybrid. Let the graph show you where the number of meeting points changes, then pin down the exact constant with algebra.

Desmos labels points with decimals, so it can’t hand you a fraction or a square root by itself. Keep those exact with algebra or by matching exact answer choices.

Check your understanding:

Which method would you start with? (1) $y=x^2-5x+4$ and $y=x+4$ , when the question asks for an exact coordinate. (2) A line and a complicated curve, when the question only asks how many times they meet.

Example: Keep every solution, then choose

Worked example

The equations $y=x^2-2x-3$ and $y=x+1$ form a system. For the solution $(x,y)$ with $x>0$ , what is the value of $y$ ?

A
$-1$
B
$0$
C
$5$
D
$4$

Step 1

Set the two expressions for y equal

At a point the graphs share, both equations give the same $y$ . So their right sides must be equal:

x^2-2x-3=x+1.

That's substitution. You've replaced $y$ with what the other equation says it equals, without rewriting anything first.

Step 2

Solve the quadratic

Move every term to one side, then factor:

\begin{aligned} x^2-2x-3&=x+1\\[1.4em] x^2-3x-4&=0\\[1.4em] (x-4)(x+1)&=0. \end{aligned}

x=4\quad\text{or}\quad x=-1.

Keep both for now. You'll use the condition $x>0$ to choose, but only once you have every point.

Step 3

Find y for both points

Plug each $x$ into the simpler equation, $y=x+1$ :

\begin{aligned} x=-1&\Longrightarrow y=0,\\[1.4em] x=4&\Longrightarrow y=5. \end{aligned}

So the system has two solutions:

(-1,0)\qquad\text{and}\qquad(4,5).

Step 4

Apply the condition and answer the question

Only $(4,5)$ has $x>0$ . The question asks for its $y$ -coordinate, so the answer is

\boxed{5},

which is C.

To be sure $(4,5)$ works, check it in both original equations:

5=4^2-2(4)-3

and

5=4+1.

Common mistake:

Stopping at the first root you find can throw away a real solution. If you had kept only $x=-1$ , you would have $(-1,0)$ and no point with $x>0$ at all. Solve every factor, find a $y$ for each root, and only then apply a condition like $x>0$ or “the greater value.”

Step 5

Check with the graph

The graph shows the same two points. You didn't need it here: algebra gave exact coordinates and made $x>0$ easy to apply. But it's a quick way to confirm you haven't missed a point.

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Click both intersections, then see which one has x greater than 0.

For more practice with this in Desmos, try Solve systems at intersections.

Solve when both equations are nonlinear

Substitution still works when neither equation is a line. All you need is one equation you can solve for a variable.

Say positive real numbers $x$ and $y$ satisfy

\begin{aligned} x^2+y^2&=13\\[1.4em] x^2-y&=7, \end{aligned}

and you want $x+y$ . Neither equation is a line, but you can get $y$ by itself from the second one:

y=x^2-7.

Substitute that into the first equation:

x^2+(x^2-7)^2=13.

Now $x^2$ shows up twice. To make the equation easier to see, call it $u$ for a moment. Let $u=x^2$ :

\begin{aligned} u+(u-7)^2&=13\\[1.4em] u+u^2-14u+49&=13\\[1.4em] u^2-13u+36&=0\\[1.4em] (u-4)(u-9)&=0. \end{aligned}

So $u=4$ or $u=9$ . Now switch back to $x^2$ . Since $x^2=4$ and $x^2=9$ each have a positive and a negative square root, keep both signs:

x=\pm2\qquad\text{or}\qquad x=\pm3.

Plug each into $y=x^2-7$ . $x=\pm2$ gives $y=-3$ , and $x=\pm3$ gives $y=2$ , so the system has four real solutions:

(-2,-3),\ (2,-3),\ (-3,2),\ (3,2).

But the problem says both numbers are positive, so only $(3,2)$ counts. That gives

x+y=3+2=\boxed{5}.

The other three points really do solve both equations. They're ruled out only because the problem said $x$ and $y$ are positive.

Check your understanding:

If the problem didn’t say $x$ and $y$ are positive, how many real solutions would the system have? Why does the actual problem keep only one?

Use a constant to get 0, 1, or 2 solutions

Some questions don't ask you to find the points. They ask how many there are, or which constant gives exactly one. Take

\begin{aligned} y&=x^2-4x+k\\[1.4em] y&=2x+3. \end{aligned}

Changing $k$ slides the parabola up or down. Depending on $k$ , the line can cross the parabola twice, touch it once, or miss it.

Set the expressions for $y$ equal:

\begin{aligned} x^2-4x+k&=2x+3\\[1.4em] x^2-6x+k-3&=0. \end{aligned}

To see the count, complete the square. $(x-3)^2$ is $x^2-6x+9$ , so $x^2-6x$ is $(x-3)^2-9$ :

\begin{aligned} x^2-6x+k-3&=0\\[1.4em] (x-3)^2-9+k-3&=0\\[1.4em] (x-3)^2&=12-k. \end{aligned}

Now the count depends only on whether $12-k$ is positive, zero, or negative:

If $k<12$ , then $12-k>0$ . A square equals a positive number in two ways, since $x-3$ can be the positive or the negative square root. So the system has two solutions.
If $k=12$ , then $12-k=0$ . The only way a square equals $0$ is $x-3=0$ , so $x=3$ and the system has exactly one solution.
If $k>12$ , then $12-k<0$ . No real number squared is negative, so the system has no real solutions.

So the constant that gives exactly one solution is

\boxed{k=12}.

At $k=12$ , the graphs touch at just one point, $(3,9)$ . That one-point touch is called tangency.

Try it yourself:

The calculator starts at k = 10, where the line crosses the parabola twice. Change k to 12, then to 14, and count the meeting points each time. Match each count to whether 12 - k is positive, zero, or negative. Click Reset before you move on.

Check your understanding:

Without finding any points, how many solutions does the system have when $k=8$ ? When $k=15$ ?

Common mistake:

Calling the graphs tangent because they look like they touch at one zoom level. They might really cross twice very close together, or miss by a hair. Use the graph to find about where two points turn into none, then prove the exact constant with algebra: the quadratic should come down to one repeated root, like $(x-3)^2=0$ .

Calculator loads as you approach

Try k = 10, 12, and 14 to see the line cross twice, touch once, then miss.

Answer what the question asks

Finding every solution and answering the question are two separate jobs. Take the solutions $(-1,0)$ and $(4,5)$ from the worked example. Depending on the question, you'd submit very different things:

If the question asks for	Submit
all the solutions	$(-1,0)$ and $(4,5)$
the positive $x$ -coordinate	$4$
the greater possible $y$ -coordinate	$5$
the value of $x+y$ when $x>0$	$9$
the number of solutions	$2$

In a word problem, also throw out any solution that breaks a condition the problem states. A negative time or length can solve the equations and still make no sense in the situation. But only throw out a solution for a reason the problem gives you, never because a number looks awkward.

Practice problems

Each calculator starts blank and saves your work. Before each problem, decide whether the graph, substitution by hand, or both will get you there.

Read two intersections from the graph

Practice problem

The graphs of

y=x^2-1.6x-0.8

and

y=0.4x+2.2

intersect at two points. What is the greater possible value of $y$ ?

Your answer

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With decimal coefficients, graphing is quickest.

Solve a line and a circle exactly

Practice problem

The system

\begin{aligned} x^2+y^2&=29\\[1.4em] y&=x+3 \end{aligned}

has two solutions. Which choice is the positive $x$ -coordinate of a solution?

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Substitution is short here. Graph afterward if you want a check.

Find the k that gives exactly one solution

Practice problem

In the system

\begin{aligned} y&=x^2-6x+k\\[1.4em] y&=4x-1, \end{aligned}

$k$ is a constant. The system has exactly one real solution. What is the value of $k$ ?

Your answer

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Find k with algebra. Graph your value afterward to see one touching point.

Finish the lesson

3 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

A solution is a point that works in both original equations.
When one variable is by itself, substitute its whole expression into the other equation.
Keep every root, find the other variable for each, and check each point in the originals.
Graph when decimal intersections or a solution count are easier to read than to calculate.
Use algebra when substitution is short, the quadratic factors, or you need to prove an exact constant.
A line and a parabola cross twice, touch once, or miss. One touching point means one repeated root.
Find every solution first, then submit only the coordinate, value, expression, or count the question asks for.

Related lesson

For more Desmos practice entering equations, finding every intersection, and adjusting the graph window, see Solve systems at intersections.

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