SHSAT: Equations true for every x, for no x, or for one x
Tell whether a linear equation has one solution, no solution or infinitely many, and find a missing constant that makes it true for every x.
How can an equation be true for every value of x, and how do you find the number that makes it so?
Short answer
Simplify both sides. If the x coefficients and the constants both match, every x works; if only the x coefficients match, no x works.
Know first: Distributing a number over parentheses, Combining like terms
Step 1 of 5
Simplify both sides, then compare them
Most equations are true for one value of x. Some are true for every value of x, and some are true for none. You can tell which by simplifying each side to the form (number)x + (number).
If both sides end up as the same expression, every x works. The equation is called an identity. If the x parts match but the constants don't, no x can ever make them equal.
| After simplifying | What happens | Example |
|---|---|---|
| a ≠ c | Exactly one solution | 5x − 3 = 3x − 5 gives x = −1 |
| a = c and b = d | Every x is a solution | 2(x − 5) + 4 = 2x − 6 |
| a = c and b ≠ d | No solution | 4x + 1 = 4x − 1 |
Compare the x coefficients first, then the constants.
Step 2 of 5
Match the coefficients, then match the constants
Worked example: Find the missing constant
For what value of m is 4(x + 3) − m = 4x + 7 true for every value of x?
- A−5
- B−4
- C5
- D19
- 1
Distribute
4(x + 3) − m = 4x + 12 − m.
- 2
Compare the x terms
Both sides have 4x. Those already match, so the rest must match too.
- 3
Compare the constants
12 − m = 7, so m = 5.
- 4
Check with any x
At x = 0: 4(3) − 5 = 7 and 4(0) + 7 = 7. At x = 1: 16 − 5 = 11 and 4 + 7 = 11.
- 5
See why the other choices are there
−5 treats −m as +m, so 12 + m = 7. −4 forgets to multiply the 3 by 4. 19 adds 12 and 7 instead of subtracting.
Answer
m = 5
Find the wrong step
A student solves 6x − 2(x − 4) = 4x + 8.
- 1
Distributes: 6x − 2x + 8 = 4x + 8.
- 2
Combines like terms: 4x + 8 = 4x + 8.
- 3
Subtracts 4x from both sides: 8 = 8.
- 4
Concludes that x = 8.
What went wrong
The x terms disappeared. The statement 8 = 8 says nothing about x, and it is always true.
The fix
Because 8 = 8 is true no matter what x is, every value of x is a solution.
Step 3 of 5
Decide how many solutions
Check yourself · Question 1
Which equation is true for every value of x?
Answer: B
The left side of B simplifies to 2x − 10 + 4 = 2x − 6, which matches the right side exactly.
Trap answer: 3(x + 2) = 3x + 2
Distributing to only the first term
- Why it looks right
- If you multiply only the x by 3, both sides read 3x + 2 and look identical.
- Why it's wrong
- The 3 multiplies the 2 as well, so the left side is 3x + 6. The constants don't match, and the equation has no solution.
- The right answer
- 2(x − 5) + 4 = 2x − 6, where both sides simplify to 2x − 6.
Check yourself · Question 2
For what value of p does 2(px + 1) = 6x + 9 have no solution?
Answer: B
The left side is 2px + 2. For no solution, the x terms must match and the constants must differ. So 2p = 6 and p = 3. Then 6x + 2 = 6x + 9, and 2 ≠ 9.
Common mistake
When the x terms cancel, I write x = 0 or x equals the leftover number.
- Why it's tempting
- You're used to every equation ending with x equal to a number.
- Do this instead
- Look at what's left. A true statement such as 8 = 8 means every x works. A false one such as 2 = 9 means no x works.
Step 4 of 5
What to remember
Remember
- Simplify each side to (number)x + (number) before deciding anything.
- Same x coefficient and same constant: every x works. Same x coefficient, different constants: no solution.
- Different x coefficients always give exactly one solution.
Step 5 of 5
Practice on a real question
Use what you just learned on this question, then check the explanation.
In your own words
In the practice question, after you expanded the left side, which part already matched and which part did you have to set equal?