Count solutions and determine constants

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Difficulty: Intermediate
Estimated time: 30 minutes
Domains: Algebra
Techniques: Solution-countUnknown-constantsCoefficient-comparisonIdentityContradiction

What you’ll learn

Spot a question that asks how many solutions an equation has, not what $x$ is.
Tell whether an equation has one solution, no solution or infinitely many, once both sides are simplified.
Find the constant that gives the number of solutions a question asks for.
Use a graph to picture the three outcomes, and use algebra to find a constant exactly.

Prerequisites

Solve linear equations0%
Distribute, combine like terms, and keep both sides equal while simplifying a one-variable equation.

Why this matters on the SAT

Make the equation behave the way the question wants

Some SAT questions don't ask you to solve for $x$ at all. Instead, they ask which constant makes a linear equation have one solution, no solution or infinitely many solutions. It all comes down to one question: once both sides are simplified, do the $x$ -terms cancel?

Solution to the example

No solution means the $x$ -terms have to cancel and leave something false behind. Start by expanding the left side:

(3k+6)x-(k+2)=15x+5.

For the $x$ -terms to cancel, the coefficients of $x$ have to match:

\begin{aligned} 3k+6&=15\\[1.4em] 3k&=9\\[1.4em] k&=3. \end{aligned}

Now check the plain numbers. At $k=3$ , the equation becomes

15x-5=15x+5.

Take $15x$ away from both sides and you're left with $-5=5$ . That's false, so no value of $x$ can work. The answer is B.

SAT example

The equation

(k+2)(3x-1)=15x+5

has no solution, where $k$ is a constant. What is the value of $k$ ?

A
$1$
B
$3$
C
$5$
D
$7$

Spot the question type before solving

These questions sound like this:

How many solutions does the equation have?
For what value of the constant does the equation have no solution?
The equation is true for all real values of $x$ .
Which condition makes the equation have exactly one solution?

You'll use the same moves as in Solve linear equations: distribute and combine like terms. The goal is different, though. Instead of finding $x$ , simplify both sides until the equation looks like

Ax+B=Cx+D.

Here $A$ and $C$ are the coefficients of $x$ , the numbers multiplying it, and $B$ and $D$ are the constants, the plain numbers. Now compare them:

After you simplify	What's left	Number of solutions
$A\ne C$	A nonzero number times $x$ equals a number	Exactly one
$A=C$ and $B\ne D$	Something false, like $4=9$	No solution
$A=C$ and $B=D$	Something true, like $4=4$	Infinitely many

Why does this work? Take $Cx+B$ away from both sides:

(A-C)x=D-B.

If $A-C$ isn't $0$ , you can divide both sides by it, and you get exactly one value of $x$ .
If $A-C$ is $0$ but $D-B$ isn't, the equation says $0$ equals some other number. That can't happen, so no $x$ works.
If both are $0$ , the equation says $0=0$ . That's true for every real value of $x$ , so every one of them solves the original equation.

Two limits. If a question only asks for the value of $x$ , solve it the usual way. And this test is for linear equations in one variable: when $x$ sits in a denominator or in a nonlinear term like $x^2$ , there are extra things to check, so use the method for that kind of equation.

Check your understanding:

After you simplify, an equation becomes $7x-4=7x-4$ . How many solutions does it have, and what’s left after you subtract $7x-4$ from both sides?

Common mistake:

This is the part that trips up a lot of students: the $x$ -terms cancel, and they answer “no solution” right away. Cancelling only tells you to look at what’s left. If it’s false, like $2=9$ , there’s no solution. If it’s true, like $2=2$ , every $x$ works, so there are infinitely many. The rule to remember: when the $x$ ’s cancel, check what’s left.

See the three outcomes as lines

You can also picture this. Graph each side of the equation as its own line. For example, $2x+1=4x-3$ becomes the lines $y=2x+1$ and $y=4x-3$ . Any point where the lines meet gives a solution, so there are three possibilities:

If the slopes are different, the lines cross once, so there's one solution.
If the slopes match but the lines sit at different heights, they're parallel and never meet, so there's no solution.
If the slopes and the heights both match, the lines lie right on top of each other, so there are infinitely many solutions.

That's the same test as the table: the slope comes from the $x$ -coefficient, and the height comes from the constant.

All four lines start out showing. Turn off the three comparison lines so only $y=2x+1$ is left, then bring them back one at a time.

Show $y=4x-3$ . The slopes are different, so the lines cross once, at $(2,5)$ . That matches the algebra: $2x+1=4x-3$ gives $x=2$ .
Hide it and show $y=2x-3$ . Same slope, different constant, so the lines are parallel.
Hide it and show $2y=4x+2$ . Divide both sides by $2$ and you get $y=2x+1$ again, so the two lines overlap completely.

Try it yourself:

Before you show each comparison line, predict how many solutions you’ll see. For the overlapping pair, turn the second equation off and on. The line changes color but doesn’t move, because both equations draw the same line.

Check your understanding:

The graph makes the three outcomes easy to see. So why isn’t it usually the best first move for finding a missing coefficient like $k$ ?

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Compare each line with y = 2x + 1: one crossing, parallel lines, then one line on top of another.

Find constants in the right order

The graph is great for seeing what's going on. When you need the exact value of a constant, though, work from the simplified equation. Take it in this order:

Simplify both sides completely. Distribute and combine like terms.
Compare the coefficients of $x$ . This tells you whether the $x$ -terms can cancel.
Compare the constants. If the $x$ -terms cancel, this tells you whether you get no solution or infinitely many.
Put your constant back in. Check that you really get the false statement, the true statement or the solvable equation you need.

Here's how that plays out in

(m+1)x+4=6x+4.

If $m=5$ , both sides become $6x+4$ , so there are infinitely many solutions. If $m$ is anything else, the coefficients of $x$ are different, so there's exactly one solution. Can any value of $m$ give no solution? No. Whenever the coefficients match, the constants already match too.

Check your understanding:

For $(p-2)x+7=3x-1$ , which values of $p$ give exactly one solution? Try to answer without solving for $x$ .

Common mistake:

Matching the coefficients of $x$ and stopping there. Matching them only makes the $x$ -terms cancel. The constants still decide between no solution and infinitely many. In the example above, $m=5$ makes the $x$ -terms cancel, but it gives infinitely many solutions, not none. So put your value back in and look at what’s left.

Example: Make both sides identical

Worked example

The equation

a(x-4)+7=5x+b

has infinitely many solutions, where $a$ and $b$ are constants. What is the value of $a+b$ ?

A
$-18$
B
$-13$
C
$-8$
D
$8$

Step 1

What “infinitely many” asks for

Infinitely many solutions means every real value of $x$ works. That only happens when both sides simplify to the same expression, so the $x$ -coefficients and the constants both have to match.

Step 2

Match the coefficients of x

Distribute $a$ on the left:

ax-4a+7=5x+b.

The coefficient of $x$ is $a$ on the left and $5$ on the right, so

a=5.

Step 3

Match the constants

The constant on the left is the whole expression $-4a+7$ , and on the right it's $b$ . Put in $a=5$ :

\begin{aligned} b&=-4a+7\\[1.4em] &=-4(5)+7\\[1.4em] &=-13. \end{aligned}

Step 4

Find a + b

The question asks for $a+b$ :

a+b=5+(-13)=\boxed{-8}.

With these values, both sides are $5x-13$ , so the equation is true for every real $x$ . The answer is C.

Common mistake:

Grabbing an answer as soon as you have $a=5$ . That’s only the first of the two matches. You still need $b$ from the constants. When you’re done, write out both full sides and make sure each one is $5x-13$ .

Practice problems

Your turn. In each problem, simplify first, then compare.

Count the solutions

Practice problem

How many solutions does the equation have?

4(2x-3)+5=8x-7

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If it helps, graph each side here and see whether the two lines overlap.

Keep exactly one solution

Practice problem

The equation

\frac32(2x+4)=kx+1

For which values of $k$ does the equation have exactly one solution?

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Work out k by hand first. Then you can graph here to check your answer.

Carry a constant to a second equation

Practice problem

A constant $c$ appears in the two equations below.

\begin{aligned} \text{I.}\quad &2(cx+5)-x=7x+10\\[1.4em] \text{II.}\quad &(c+1)x-3=5x+2 \end{aligned}

If equation I is true for all values of $x$ , how many solutions does equation II have?

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Find c from equation I first. Then you can graph equation II here to check.

Match both parts

Practice problem

The equation

p(2x-3)+q=10x-7

has infinitely many solutions, where $p$ and $q$ are constants. What is the value of $p+q$ ?

Your answer

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Find p and q by hand. Then graph both sides here to see them land on the same line.

Finish the lesson

4 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

Simplify both sides, then compare the coefficients of $x$ and the constants.
Different $x$ -coefficients give exactly one solution.
Equal $x$ -coefficients with different constants give no solution.
Equal $x$ -coefficients with equal constants give infinitely many solutions.
To find a constant, match the $x$ -coefficients first, check the constants second, and put your value back in.
A graph shows crossing, parallel or overlapping lines, but algebra is usually the faster, exact way to pin down a constant.

Next lesson

Find slope and rate of change

Find and interpret slope from points, tables, graphs, and equations.

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Practice

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491 SAT questions use what this lesson teaches. Practice a few in a study session at the difficulty you choose.

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