When the x cancels, the equation has stopped being a question about x and started being a statement about the two constants.
In the given equation, a is a constant.
3(4x + 7) = ax − 12
For what value of a does the equation have no solution?
Distribute first: 12x + 21 = ax − 12. The two sides can only fail to meet if they climb at the same rate, so a = 12, and the equation becomes 12x + 21 = 12x − 12, that is 21 = −12, which is false for every x. C is 4, the coefficient still trapped inside the bracket — the value you get by comparing before distributing. B is 3, the multiplier outside the bracket. A is −12, the constant on the right, which is the number you would match if the question said infinitely many solutions — and even then 21 ≠ −12, so no value of a can do that here.
1
Solve as usual, and watch for the moment the x disappears.
Nothing new happens until the last step. Subtract the x-terms and one of three things is left.
2x + 5 = 3x + 9 → −4 = x
different x-terms, so exactly one solution
2x + 5 = 2x + 9 → 5 = 9
x gone, and what is left is false → no solution
2(2x + 2) = 4x + 4 → 4 = 4
x gone, and what is left is true → infinitely many
Last line reads
Meaning
Number of solutions
x = a number
one value works
exactly one
a false number sentence
no value works
zero — “no solution”
a true number sentence
every value works
infinitely many
“5 = 9” is not an error you made. It is the answer: you have proved that no number x can satisfy the equation. Likewise “4 = 4” is not a dead end — it says the equation was never a restriction on x at all.
2
Tidy both sides to mx + b, then compare two numbers.
No solution: the same coefficient of x, different constants. Infinitely many: the same coefficient of x and the same constant. Anything else: exactly one solution.
Left side
Right side
Compare
Solutions
7x − 2
3x + 6
7 ≠ 3
exactly one
7x − 2
7x + 6
7 = 7, −2 ≠ 6
none
7x − 2
7x − 2
identical
infinitely many
same multiple of x, different constant → nonethe very same expression → infinitely many
Put in words: no solution means the two sides are the same multiple of x with different constants added, so one side always sits a fixed distance from the other. Infinitely many means the two sides are the same expression wearing different clothes.
3
“For what value of a does the equation have no solution?”
This is the opening problem. Distribute, then match the x-coefficients and check that the constants disagree.
3(4x + 7) = ax − 12
the given equation
12x + 21 = ax − 12
3 · 4x = 12x and 3 · 7 = 21
coefficients: 12 and a
set them equal: a = 12
constants: 21 and −12
21 ≠ −12 ✓ so it really is no solution
check: 12x + 21 = 12x − 12 → 21 = −12
false for every x ✓
a = 12
Match coefficients only after you distribute
Comparing a with the 4 inside 3(4x + 7) gives a = 4, which is why 4 is on the answer list. With a = 4 the equation is 12x + 21 = 4x − 12, an ordinary equation with the single solution x = −338. Multiply out first, every time.
4
“…infinitely many solutions?” means make the two sides identical.
Now the constants have to match as well, so there is a second equation to solve.
−2(5x − 7) = ax + k
a and k are constants. For what values of a and k does the equation have infinitely many solutions?
−10x + 14 = ax + k
−2 · 5x = −10x and −2 · (−7) = +14
a = −10
the x-terms must agree
k = 14
and so must the constants
test x = 1: −2(5 − 7) = 4 and −10 + 14 = 4 ✓
test x = 0: 14 = 14 ✓
When the parameter sits inside a fraction or a bracket, multiply it out before you compare. If (34)(8x + c) = 6x − 15 is to have infinitely many solutions, the left side is 6x + (34)c, so (34)c = −15 and c = −20 — not −15.
Two different questions live in the same sentence. “No solution” fixes one number and requires the other pair to disagree; “infinitely many” fixes both. If a problem gives you only one constant to find and asks for infinitely many, the other pair must already match — check that it does.
5
The trap: solving for x instead of comparing coefficients.
Here is the student who treats it as an ordinary equation and grinds on.
12x + 21 = ax − 12
collect everything
12x − ax = −33
x-terms left, constants right
(12 − a)x = −33
factor the x out
x = −3312 − a
and now there is nothing numerical to report
The answer is hiding in the step you skipped
Dividing by 12 − a quietly assumed 12 − a ≠ 0. That assumption is the whole question: when 12 − a = 0, the line (12 − a)x = −33 reads 0 = −33, which no x satisfies. So a = 12, and the student who is hunting for a numeric x never sees it. Whenever a coefficient contains an unknown constant, stop at (something)x = (something) and ask what happens if the coefficient is zero.
The same reading handles the bare form. In (m − 7)x = 20, the left side collapses to 0 when m = 7 and 0 = 20 is false, so m = 7 gives no solution; every other m gives exactly one.
6
In the xy-plane the two cases look completely different.
Graph the left side and the right side as two functions. A solution of the equation is a point where the graphs meet.
Where the calculator earns its keep
Type the left side and the right side into Desmos as two separate functions. Two parallel lines with no intersection means no solution; a single line, because the second graph lands exactly on the first, means infinitely many. With a constant in the equation, define it with a slider — drag a until the crossing point flies off the screen and you are looking at the value the question wants.
✓
Check: put the value back and read the last line.
a = 1212x + 21 = 12x − 1221 = −12, false → none ✓
a = −10, k = 14−10x + 14 = −10x + 14identical → infinitely many ✓
m = 70 · x = 200 = 20, false → none ✓
Remember the flow
Distribute · Compare the x-coefficients · Then compare the constants
Never divide by a coefficient that contains the constant you are looking for.
Now you try
Practice problem
In the given equation, c is a constant.
23(6x + c) = 4x − 22
If the equation has infinitely many solutions, what is the value of c?
Distribute the 23: the left side is 4x + (23)c. The x-terms already agree, so the equation is true for every x exactly when (23)c = −22, giving c = −22 · 32 = −33. Check: (23)(6x − 33) = 4x − 22 ✓ B is −22, the constant copied straight across as though the 23 only touched the x-term. C is −443, from multiplying by 23 instead of dividing. D is 33, the right size with the sign lost along the way.
Hint: get the left side into the form (number)x + (number) before you compare anything. Then ask the two questions in order — do the x-terms match, and do the constants have to match too?
