Equations with No or Infinitely Many Solutions, SAT math lesson

Two phone plans both charge 15 dollars a month plus 2 dollars per gigabyte. Ask when they cost the same and you get $15 + 2g = 2g + 15$ , which holds for every amount of data. Raise the base price of one plan to 20 dollars and the question becomes $20 + 2g = 2g + 15$ , which never holds. In both equations the variable cancels when you solve, and the small statement left behind decides the answer.

See also Solving Equations and Checking Solutions in Foundations for Calculus.

Key terms

Coefficient

Definition In a term that is the product of a number and a variable, such as $7x$ , the number is the coefficient of the variable. A term written $x$ has coefficient $1$ , and a term written $-x$ has coefficient $-1$ . A letter that stands for a fixed number, such as $a$ or $k$ in $kx$ , can also be a coefficient. When you compare the two sides of an equation, read each coefficient only after that side is expanded and like terms are combined.

In plain words The coefficient is the number that multiplies $x$ . It tells you how many copies of $x$ a side contains.

Example In $7x - 4 = -x + 9$ the coefficient of $x$ is $7$ on the left and $-1$ on the right. In $3(2x - 1)$ the coefficient of $x$ is not $3$ . Expanding gives $6x - 3$ , so the coefficient is $6$ .

Constant term

Definition A constant term is a term that contains no variable, so its value does not depend on $x$ . After expanding and combining like terms, every linear equation in $x$ can be written as

ax + b = cx + d,

where $a$ and $c$ are the coefficients of $x$ and $b$ and $d$ are the constant terms.

In plain words The constant is the part of a side that stays the same no matter what number you put in for $x$ .

Example The left side of $4(x + 2) - 3 = 4x + 5$ expands to $4x + 8 - 3$ , which is $4x + 5$ . Its constant term is $5$ , the combination of $8$ and $-3$ . So both sides have coefficient $4$ and constant term $5$ .

Identity

Definition An equation in $x$ is an identity when it is true for every value of $x$ in its domain, the set of values for which both sides are defined. When equivalent steps make the variable disappear and leave a true statement without a variable, such as $6 = 6$ or $0 = 0$ , the original equation is an identity. A linear equation $ax + b = cx + d$ is an identity exactly when $a = c$ and $b = d$ .

In plain words Both sides are the same expression written two ways. Whatever number you try, the two sides match.

Example Solve $2(x + 3) = 2x + 6$ . Expanding gives $2x + 6 = 2x + 6$ , and subtracting $2x$ from both sides leaves

6 = 6,

which is always true. Every real number is a solution. A spot check agrees. At $x = -5$ both sides equal $-4$ , and at $x = 10$ both equal $26$ .

Infinitely many solutions

Definition An equation has infinitely many solutions when its solution set contains infinitely many numbers. The solution set of an identity is every value in the original domain. For a linear identity that is every real number. If the original equation excludes some values, for instance because they make a denominator zero, those values stay excluded even when simplifying hides the denominator.

In plain words A true leftover statement means every number that was allowed in the first place works. It does not let back in a number that the original equation could not use.

Example Consider

\frac{3x + 6}{x + 2} = 3.

The value $x = -2$ makes the denominator zero, so the domain is every real number except $-2$ . For every other $x$ the factor $x + 2$ cancels,

\frac{3x + 6}{x + 2} = \frac{3(x + 2)}{x + 2} = 3,

and the equation reduces to the true statement $3 = 3$ . So it is an identity on its domain, and its solution set is all real numbers except $-2$ . That is still infinitely many solutions, but $x = -2$ is not one of them, because the left side is undefined there.

Contradiction

Definition An equation in $x$ is a contradiction when no value of $x$ in its domain makes it true. When equivalent steps make the variable disappear and leave a false statement without a variable, such as $6 = 7$ or $0 = 5$ , the original equation is a contradiction. A linear equation $ax + b = cx + d$ is a contradiction exactly when $a = c$ and $b \ne d$ .

In plain words The two sides grow at the same rate but start from different places, so they are always the same distance apart and never meet.

Example Solve $3(x - 2) + 12 = 3x + 7$ . The left side simplifies to $3x + 6$ . Subtracting $3x$ from both sides leaves

6 = 7,

which is false. Trying numbers shows why. At $x = 0$ the sides are $6$ and $7$ , and at $x = 10$ they are $36$ and $37$ . The left side is always exactly $1$ less than the right.

No solution

Definition An equation has no solution when its solution set is empty, written $\varnothing$ . Every contradiction has no solution. When the variable cancels, the leftover constant statement decides the case. A true statement means every allowed value is a solution, and a false statement means none is. A canceled variable never means $x = 0$ by itself.

In plain words If the $x$ terms vanish, do not write $x = 0$ . Look at what is left. A false statement says no number works, zero included.

Example Solve $5x - 2 = 5x + 4$ . A common slip is to cancel the $5x$ terms and write $x = 0$ . Test it. At $x = 0$ the sides are $-2$ and $4$ , which are not equal. The correct reading is that canceling leaves $-2 = 4$ , which is false, so the equation has no solution.

Compare $5x - 2 = 3x - 2$ . Here the variable does not cancel. Subtracting $3x$ and adding $2$ gives $2x = 0$ , so $x = 0$ really is the one solution, and both sides equal $-2$ there.

A graph shows the same thing. In the Desmos graphing calculator, type $y = 5x - 2$ and $y = 5x + 4$ . The two lines have the same slope and different $y$ -intercepts, so they are parallel and never cross. For an identity such as $y = 2(x + 3)$ and $y = 2x + 6$ , the second line lies exactly on top of the first.

Parameter cases

Definition A parameter is a letter, such as $k$ , that stands for a fixed but unknown number. After both sides are simplified to $ax + b = cx + d$ , subtracting $cx$ and $b$ from both sides gives

(a - c)x = d - b.

So the equation has

exactly one solution when $a \ne c$ , namely $x = \frac{d - b}{a - c}$ ,
infinitely many solutions, every real number, when $a = c$ and $b = d$ ,
no solution when $a = c$ and $b \ne d$ .

Dividing by $a - c$ is allowed only when $a - c \ne 0$ , which is why the case $a = c$ has to be handled separately.

In plain words Compare the coefficients first. Different coefficients give one solution. Equal coefficients send you to the constants, where equal means infinitely many and different means none.

Example In $4(x - k) = 4x - 12$ , the number $k$ is a constant. Expanding gives $4x - 4k = 4x - 12$ . The coefficients are $4$ and $4$ for every $k$ , so this equation never has exactly one solution. The constants match when $-4k = -12$ , that is when $k = 3$ . So the equation has infinitely many solutions for $k = 3$ and no solution for every other $k$ .

With two parameters, take $kx + 5 = 2x + m$ . Rearranging gives $(k - 2)x = m - 5$ , and all cases fit in one table.

Condition	Number of solutions
$k \ne 2$	exactly one, $x = \frac{m - 5}{k - 2}$
$k = 2$ and $m = 5$	infinitely many
$k = 2$ and $m \ne 5$	none

On the SAT these questions usually say that a letter is a constant and give one case, such as “the equation has no solution”, then ask for the constant. Match the coefficients, then make the constants equal or different as the case requires. A graph shows the same split. In Desmos, graph $y = kx + 5$ with a slider for $k$ next to $y = 2x + 7$ , which is the case $m = 7$ . The lines cross at one point for every $k$ except $k = 2$ , where they are parallel and the equation has no solution.

Check yourself

Which of the following equations has no solution? (A) $4(x - 2) = 4x - 8$ (B) $4(x - 2) = 4x - 2$ (C) $4(x - 2) = 2x - 8$ (D) $4(x - 2) = -4x + 8$

Show the answer

The correct choice is (B). The left side expands to $4x - 8$ in every choice. In (B) canceling $4x$ leaves $-8 = -2$ , which is false, so there is no solution. Choice (A) leaves $-8 = -8$ , an identity with infinitely many solutions. Choice (C) gives $2x = 0$ , so $x = 0$ is one solution. Choice (D) gives $8x = 16$ , so $x = 2$ is one solution.

In the equation $5(x + k) - 3 = 5x + 17$ , the number $k$ is a constant. If the equation has infinitely many solutions, what is the value of $k$ ?

Show the answer

The value is $4$ . The left side is $5x + 5k - 3$ , and its coefficient already matches the $5$ on the right. For infinitely many solutions the constants must match too, so $5k - 3 = 17$ , which gives $5k = 20$ and $k = 4$ . Then both sides are $5x + 17$ .

In the equation $ax - 6 = 3(x - 2)$ , the number $a$ is a constant. Which statement is true? (A) The equation has no solution when $a = 3$ . (B) The equation has infinitely many solutions for every value of $a$ . (C) The equation has infinitely many solutions when $a = 3$ and exactly one solution, $x = 0$ , otherwise. (D) The equation has no solution when $a \ne 3$ .

Show the answer

The correct choice is (C). The right side is $3x - 6$ , so the equation becomes $(a - 3)x = 0$ . When $a = 3$ this reads $0 = 0$ , true for every $x$ , so (A) fails. When $a \ne 3$ you may divide by $a - 3$ and get $x = 0$ , exactly one solution, so (B) and (D) fail. Here $x = 0$ is a real solution because the variable survived with a nonzero coefficient, not because it canceled.

Equations with No or Infinitely Many Solutions

Theory, part 17

Practice, part 18

Key terms

Coefficient

Constant term

Identity

Infinitely many solutions

Contradiction

No solution

Parameter cases

Check yourself

How this lesson connects

Builds on

Used later in