Systems with One, No or Infinite Solutions, SAT math lesson

Draw one line on a sheet of paper and lay a ruler across it. The ruler either crosses the line at one point, runs beside it without ever touching, or covers it exactly. Two linear equations behave the same way, and the SAT often asks which of the three pictures you have without asking you to solve anything. The trick is to read the answer from the coefficients, carefully enough that you never confuse the last two pictures.

See also Linear Systems in Two and Three Variables in Precalculus.

Key terms

Number of solutions

Definition Consider a system of two linear equations in $x$ and $y$ ,

\begin{aligned} a_1 x + b_1 y &= c_1 \\[0.4em] a_2 x + b_2 y &= c_2, \end{aligned}

where all numbers are real and in each equation the coefficients $a$ and $b$ are not both zero, so each equation describes a line. A solution is an ordered pair $(x, y)$ that makes both equations true. Such a system has exactly one solution, no solution or infinitely many solutions, and no other count is possible. It has exactly one solution exactly when

a_1 b_2 - a_2 b_1 \ne 0,

which for nonvertical lines says that the two slopes are different.

In plain words Two different slopes mean the lines cross, and they cross once. Equal slopes leave only two options, never meeting or being the same line.

Example In the system $2x + 3y = 12$ and $x - y = 1$ , the check gives $2(-1) - 1 \cdot 3 = -5$ , which is not zero, so there is exactly one solution. Solving confirms it. The second equation gives $x = y + 1$ , and then $2(y + 1) + 3y = 12$ gives $y = 2$ and $x = 3$ . The lines have slopes $-\frac{2}{3}$ and $1$ and meet at $(3, 2)$ .

Consistent system

Definition A system is consistent when it has at least one solution. A consistent linear system of two equations in two variables has either exactly one solution, when the lines cross, or infinitely many, when the lines coincide.

In plain words Consistent means the equations can all be true at once. The lines share at least one point.

Example The system $y = 2x + 1$ and $y = -x + 7$ is consistent. Setting the right sides equal gives $2x + 1 = -x + 7$ , so $x = 2$ and $y = 5$ . Both equations hold at $(2, 5)$ , since $2 \cdot 2 + 1 = 5$ and $-2 + 7 = 5$ .

Inconsistent system

Definition A system is inconsistent when it has no solution. A system of two linear equations in two variables is inconsistent exactly when its lines are parallel and distinct, that is when they have the same slope (or are both vertical) but are different lines. Elimination then removes both variables at once and leaves a false statement such as $0 = 5$ .

In plain words The two equations contradict each other. The lines run side by side forever and never share a point.

Example Take $3x - y = 4$ and $6x - 2y = 3$ . Multiply the first equation by $2$ to line up the left sides.

\begin{aligned} 6x - 2y &= 8 \\[0.4em] 6x - 2y &= 3 \end{aligned}

Subtracting gives $0 = 5$ , which is false for every pair $(x, y)$ . In slope form the lines are $y = 3x - 4$ and $y = 3x - 1.5$ , same slope, different intercepts.

Coincident lines

Definition Two linear equations describe coincident lines when they have exactly the same solution set, that is when one equation is a nonzero multiple of the other, both sides included. The system then has infinitely many solutions, namely every point of that one line. Elimination removes both variables and leaves a true statement such as $0 = 0$ .

Matching coefficient ratios are not enough to tell coincident lines from parallel ones. The ratios $\frac{a_1}{a_2} = \frac{b_1}{b_2}$ only say that the slopes agree. The constants decide the rest. To avoid dividing by a coefficient that may be zero, compare cross products instead of ratios. The lines coincide exactly when

\begin{gathered} a_1 b_2 = a_2 b_1, \quad a_1 c_2 = a_2 c_1, \\[0.4em] b_1 c_2 = b_2 c_1, \end{gathered}

and they are parallel and distinct when the first equation holds but one of the other two fails.

In plain words Coincident lines are one line written twice. Every point on it solves the system, but a point off the line still does not.

Example For $2x + 5y = 7$ and $6x + 15y = 21$ the second equation is $3$ times the first, constants included, so the lines coincide. The pair $(1, 1)$ works in both, and so does $(-4, 3)$ . The pair $(0, 0)$ is not a solution, so infinitely many solutions does not mean every pair. Now change only the constant to get $6x + 15y = 20$ . The coefficient ratios are still $\frac{6}{2} = \frac{15}{5} = 3$ , but $3 \cdot 7 = 21 \ne 20$ , so the lines are parallel and distinct and the system has no solution.

Ratios can also break down when a coefficient is $0$ . In $y = 4$ and $2y = 8$ the ratio $\frac{a_1}{a_2}$ would be $\frac{0}{0}$ , which is undefined. The cross products give $0 \cdot 2 = 0 \cdot 1$ , $0 \cdot 8 = 0 \cdot 4$ and $1 \cdot 8 = 2 \cdot 4$ , so the lines coincide.

Parameter

Definition A parameter is a letter in a system, such as $k$ or $a$ , that stands for a fixed real number which is not stated. It is not one of the unknowns. For each value of the parameter you get a separate system in $x$ and $y$ , and a question may ask which value gives a particular number of solutions.

In plain words The parameter is a dial. Turn it to a number and you have an ordinary system to classify.

Example Find $k$ so that $2x + 5y = 7$ and $6x + ky = 21$ have infinitely many solutions. The $x$ coefficient and the constant are both multiplied by $3$ , so the whole second equation must be $3$ times the first, which forces $k = 15$ . For any other $k$ the slopes differ and there is exactly one solution. No value of $k$ gives no solution here, since the constants already match the multiple $3$ .

Parameter cases

Definition Classifying a system with a parameter means splitting the possible values of the parameter into cases and giving the exact number of solutions in each case. Before you divide by an expression that contains the parameter, find the values that make it zero and treat them as separate cases. A step that is valid for most values, such as dividing by $k^2 - 4$ , can be invalid for a few, and those few are usually the interesting ones.

In plain words Find the values where the usual method breaks, deal with them on their own, and then solve the general case.

Example Classify $kx + 2y = 4$ and $2x + ky = 4$ for every real $k$ . Multiply the first equation by $k$ and the second by $2$ , then subtract to remove $y$ .

\begin{aligned} (k^2 - 4)x &= 4k - 8 \\[0.4em] (k + 2)(k - 2)x &= 4(k - 2) \end{aligned}

The coefficient of $x$ is zero when $k = 2$ or $k = -2$ , so look at those first.

Case	System	Result
$k = 2$	$2x + 2y = 4$ twice	infinitely many solutions
$k = -2$	$-2x + 2y = 4$ and $2x - 2y = 4$	no solution
$k \ne \pm 2$	divide by $(k + 2)(k - 2)$	exactly one solution

For $k = -2$ , adding the two equations gives $0 = 8$ , a contradiction. In the last case

x = \frac{4}{k + 2}, \qquad y = \frac{4}{k + 2}.

Dividing at the start would have hidden both special cases. The multiplication needs a second look too. Multiplying the first equation by $k$ cannot be undone when $k = 0$ , so the subtraction only tells you which value $x$ must take. That is still enough here, because the first equation then fixes $y$ through $2y = 4 - kx$ . At $k = 0$ the formulas give $(2, 2)$ , and both $2y = 4$ and $2x = 4$ hold.

Check yourself

In the system $4x - 2y = 6$ and $y = 2x + c$ , where $c$ is a constant, for which value of $c$ does the system have infinitely many solutions? (A) $-3$ (B) $3$ (C) $-6$ (D) $6$

Show the answer

The answer is (A). Solving the first equation for $y$ gives $y = 2x - 3$ . The second line has the same slope $2$ , so the lines coincide exactly when $c = -3$ . For any other $c$ they are parallel and distinct and the system has no solution. Each wrong choice gives a parallel line with a different intercept, so the system would have no solution. Choice (B) comes from moving $4x$ across but keeping the sign of $6$ , which gives $2y = 4x + 6$ and $y = 2x + 3$ . Choice (C) comes from dividing $4x$ by $2$ but not the $6$ , which turns $2y = 4x - 6$ into $y = 2x - 6$ . Choice (D) copies the constant $6$ from the first equation without rewriting it in slope form at all.

The system $3x + ky = 8$ and $12x + 20y = 40$ has no solution, where $k$ is a constant. What is the value of $k$ ?

Show the answer

The answer is $5$ . No solution needs equal slopes, so the second equation’s coefficients must be the same multiple of the first. Since $12 = 4 \cdot 3$ , the condition $4k = 20$ gives $k = 5$ . The constants do not follow that multiple, because $4 \cdot 8 = 32 \ne 40$ , so the lines are parallel and distinct rather than coincident.

In the system $(a - 1)x + y = 3$ and $2x + y = a$ , where $a$ is a constant, which statement is true? (A) The system has exactly one solution for every value of $a$ . (B) The system has no solution when $a = 3$ . (C) The system has infinitely many solutions when $a = 3$ and exactly one solution for every other value of $a$ . (D) The system has no solution when $a = 1$ .

Show the answer

The answer is (C). Subtracting the second equation from the first gives $(a - 3)x = 3 - a$ . When $a = 3$ this reads $0 = 0$ , and indeed both equations become $2x + y = 3$ , so the lines coincide. For $a \ne 3$ you may divide by $a - 3$ and get $x = -1$ and then $y = a + 2$ , one solution. Choice (A) misses the value $a = 3$ , where the coefficient $a - 3$ is zero and you cannot divide by it. Choice (B) names the right value but assumes that equal slopes mean no solution without checking the constants, and choice (D) fails because at $a = 1$ the first line is $y = 3$ , which still crosses $2x + y = 1$ at $(-1, 3)$ .

Systems with One, No or Infinite Solutions

Theory, part 33

Practice, part 34

Key terms

Number of solutions

Consistent system

Inconsistent system

Coincident lines

Parameter

Parameter cases

Check yourself

How this lesson connects

Builds on

Used later in