Solve linear equations in one variable and classify them as having one, no, or infinitely many solutions

Learning goal: Solve linear equations in one variable, give examples with one solution, infinitely many solutions or no solutions, and show which of these possibilities is the case by successively transforming the given equation into simpler forms until an equivalent equation of the form x = a, a = a, or a = b results, where a and b are different numbers.

Linear equations in one variable. A linear equation in one variable is one that can be rearranged into the form ax + b = cx + d, where a, b, c and d are numbers and the variable appears only to the first power. A solution is a number that makes the two sides equal when it is substituted for x. Two equations are equivalent when they have exactly the same solutions.

Moves that keep equations equivalent. Adding the same quantity to both sides, subtracting the same quantity from both sides, multiplying both sides by the same nonzero number, and dividing both sides by the same nonzero number all produce an equivalent equation, because each move can be reversed. Replacing one side by an equivalent expression, such as combining like terms or distributing, is also allowed because it changes the form and not the value. Multiplying both sides by 0 is not allowed: it turns every equation into 0 = 0 and destroys the information.

The three possibilities. When you keep simplifying, one of three things happens. You reach x = a, a single number: one solution. You reach a = a, a statement that is true no matter what x is, because x has vanished from both sides: infinitely many solutions, every number works. You reach a = b with a and b different numbers, a statement that is false no matter what x is: no solutions. Successively transforming the given equation into simpler forms is the method; the final form tells you which possibility is the case.

A fully worked model: one solution. Solve 5x + 7 = 2x + 22. Subtract 2x from both sides: 3x + 7 = 22. Subtract 7 from both sides: 3x = 15. Divide both sides by 3: x = 5. The final form is x = a with a = 5, so there is exactly one solution. Check by substitution in the original: 5(5) + 7 = 32 and 2(5) + 22 = 32. Both sides agree.

A fully worked model: infinitely many solutions. Solve 4(x + 3) = 4x + 12. Distribute on the left: 4x + 12 = 4x + 12. Subtract 4x from both sides: 12 = 12. The variable has disappeared and the statement is true. This is the form a = a, so every number is a solution. Check two very different values in the original: x = 0 gives 4(3) = 12 and 0 + 12 = 12; x = −2.5 gives 4(0.5) = 2 and 4(−2.5) + 12 = −10 + 12 = 2. Such an equation is called an identity: the two sides are the same expression written two ways.

A fully worked model: no solutions. Solve 3x + 8 = 3x − 1. Subtract 3x from both sides: 8 = −1. This is the form a = b with a and b different numbers, and it is false for every x. There are no solutions. The reason is visible in the original: 3x + 8 is always 9 more than 3x − 1, so the two can never be equal.

Producing your own examples. To make an equation with infinitely many solutions, write an expression and an equivalent form of it on the other side: 2(3x − 4) = 6x − 8. To make one with no solutions, keep the same coefficient of x on both sides but use different constants: 6x − 8 = 6x + 1. To make one with a single solution, use different coefficients of x: 6x − 8 = 2x + 4, which simplifies to 4x = 12 and x = 3; check, 18 − 8 = 10 and 6 + 4 = 10. The rule behind this: in ax + b = cx + d, if a and c differ there is one solution; if a = c and b = d every number works; if a = c but b and d differ, nothing works.

Misconceptions to avoid. Reaching 0 = 0 does not mean the solution is x = 0; it means x dropped out and every value works. Reaching 8 = −1 does not mean the solutions are 8 and −1; it means there are none. When subtracting 3x from both sides leaves no x, students often assume they made an error and start over; the disappearance is the result, not a mistake. Finally, a check must go back to the original equation, not to a later line, because an arithmetic slip in the middle would make a later line agree with a wrong answer.

Worked one solution: 7x − 9 = 4x + 12. Subtract 4x: 3x − 9 = 12. Add 9: 3x = 21. Divide by 3: x = 7. Form x = a. Check in the original: 7(7) − 9 = 49 − 9 = 40 and 4(7) + 12 = 28 + 12 = 40.

Worked infinitely many: 2(4x − 5) = 8x − 10. Distribute: 8x − 10 = 8x − 10. Subtract 8x: −10 = −10. Form a = a, so every number is a solution. Check x = 1: 2(4 − 5) = −2 and 8 − 10 = −2. Check x = 1/2: 2(2 − 5) = −6 and 4 − 10 = −6.

Worked no solutions: 5x + 3 = 5(x + 2). Distribute: 5x + 3 = 5x + 10. Subtract 5x: 3 = 10. Form a = b with different numbers, so there are no solutions. Compare with the previous example: the only change from an identity to an impossible equation is the constant, because the coefficients of x are the same on both sides.

Worked one solution with a negative coefficient: −3(x − 4) = 2x + 2. Distribute: −3x + 12 = 2x + 2. Add 3x to both sides: 12 = 5x + 2. Subtract 2: 10 = 5x. Divide by 5: x = 2. Check: −3(2 − 4) = −3(−2) = 6 and 2(2) + 2 = 6.

Worked family of examples: Consider 9x + c = 3(3x + 4), which is 9x + c = 9x + 12. Subtracting 9x leaves c = 12. If c is 12, the form is a = a and there are infinitely many solutions; for any other value of c the form is a = b and there are none. This equation can never have exactly one solution, because the coefficient of x is 9 on both sides. By contrast 9x + 5 = kx + 5 becomes (9 − k)x = 0: one solution, x = 0, whenever k is not 9, and infinitely many when k = 9.

Representation task: Make a three-column table headed one solution, infinitely many solutions, no solutions. Under each heading write two original equations, each containing parentheses on at least one side. For each equation, show the successive transformations line by line until you reach x = a, a = a, or a = b, and state the form reached. For the one-solution equations, check the value in the original equation with the arithmetic written out.

Independent performance: Write the equation 4(2x − 3) + p = 8x + q with blanks p and q. Explain which values of p and q give infinitely many solutions, which give none, and why no choice gives exactly one solution. Then change one coefficient so that the equation does have exactly one solution, solve it, and check. Success criteria: each classification is justified by the simplified form, not by guessing; the explanation states the general rule about equal or different coefficients of x; the check substitutes into the original equation; no claim that 0 = 0 means x = 0 or that a = b names two solutions.