Two-Step Inequalities Practice Test | Free Algebra Questions

After the test · two-stage inequality control

A two-step inequality is solved by removing the constant first and then undoing the coefficient or divisor attached to the variable

The structure is deliberately narrow: one step clears addition or subtraction, and a second step clears multiplication or division. The inequality direction changes only if that second operation involves multiplying or dividing by a negative number.

ax + b < cax − b ≥ cx/a + b ≤ c Positive coefficientNegative coefficientBoundary check

Keep the two jobs separate: Step 1 isolates the variable term. Step 2 isolates the variable itself.

Stage 1 · constant Undo the addition or subtraction outside the variable term.

Example: 3x + 5 < 20 → 3x < 15.

→

Stage 2 · coefficient Undo the multiplication or division attached to x.

3x < 15 → x < 5. Since 3 is positive, the sign stays the same.

Two-step equation families

The same two-stage logic appears in several common forms

Addition form

4x + 7 ≤ 23

Subtract 7, then divide by 4.

Subtraction form

5x − 6 > 14

Add 6, then divide by 5.

Division form

x/3 + 2 ≥ 8

Subtract 2, then multiply by 3.

Negative coefficient

−2x + 5 < 13

Subtract 5, then divide by −2 and reverse the sign.

Two-step trace · 4x − 7 ≤ 13

START

4x − 7 ≤ 13 The variable term is 4x; −7 is the outside constant.

STEP 1

4x ≤ 20 Add 7 to both sides. The sign does not change.

STEP 2

x ≤ 5 Divide by positive 4. The inequality direction stays the same.

CHECK

x = 5 gives 4(5) − 7 = 13 The boundary value works, so the inclusive symbol ≤ is consistent.

Second-step decision gate

The first step never flips the sign when it is ordinary addition or subtraction; the second step may

Positive coefficient 3x + 4 > 16
3x > 12
x > 4

Divide by +3. Keep the direction.

STEP 2
CHECK SIGN

Negative coefficient −3x + 4 > 16
−3x > 12
x < −4

Divide by −3. Reverse > to <.

Negative-coefficient route

Do the same two steps, but reserve the sign reversal for the moment you divide by the negative coefficient

Example: −2x + 5 < 13.

Original −2x + 5 < 13

Two operations affect x.

Step 1 −2x < 8

Subtract 5. Keep <.

Step 2 x > −4

Divide by −2 and reverse the sign.

Quick test x = 0 → 5 < 13

0 is greater than −4, so the direction is plausible.

Fraction and decimal two-step lane

Decimal and fractional coefficients still use exactly two inverse operations

Decimal coefficient

0.5x + 2 > 7 → 0.5x > 5 → x > 10

Positive division preserves the direction.

Fractional coefficient

(2/3)x − 1 ≤ 5 → (2/3)x ≤ 6 → x ≤ 9

Multiply by positive reciprocal 3/2.

Negative decimal

−0.25x + 3 ≥ 5 → −0.25x ≥ 2 → x ≤ −8

Division by −0.25 reverses ≥ to ≤.

Solution direction

The final inequality determines both the endpoint type and the shading direction

x ≤ 5

5

Closed endpoint because 5 is included; shade toward smaller values.

x > −4

−4

Open endpoint because −4 is excluded; shade toward larger values.

Two-step context model

A fixed amount plus a repeated amount naturally creates a two-step inequality

Budget limit

$50

Fixed fee

$8

Cost per item

$6

Number of items

x

Inequality

8 + 6x ≤ 50

Solve

x ≤ 7

Boundary verification

Test the boundary in the original inequality, not only in the simplified result

8 + 6x ≤ 50
x ≤ 7

→

x = 7
8 + 6(7)

→

50 ≤ 50
TRUE ✓

Two-step error map

The most common mistakes happen when the two inverse operations are performed in the wrong order or the second-step sign rule is missed

Wrong first step

4x + 7 ≤ 23 → x + 7 ≤ 23/4 4x ≤ 16 → x ≤ 4

Remove the outside constant before dividing by the coefficient.

Flip too early

−2x + 5 < 13 → −2x > 8 −2x < 8

Subtracting 5 does not reverse the inequality. The flip happens only when dividing by −2.

Forgot second-step flip

−2x < 8 → x < −4 x > −4

Division by a negative reverses the sign.

Boundary mismatch

x ≤ 5 shown with an open endpoint Use a closed endpoint at 5.

The equality part of ≤ includes the boundary.

Two-stage diagnostics

Sort missed questions by which of the two steps or final interpretation caused the error

Step 1 Was the constant removed with the correct inverse operation?

Step 2 Was the coefficient or divisor undone correctly?

Sign reversal Was the inequality flipped only when Step 2 used a negative factor?

Boundary & direction Did the endpoint and shading match the final inequality?