Notes · Unit 5
Linear Equations and Inequalities
Thirteen short slides on linear equations, absolute value, radical equations, linear inequalities and compound inequalities.
Press Next (or the right arrow key) to move on. On most slides, each press reveals one more idea.
Linear equationsAbsolute valueRadical equationsInequalitiesCompound inequalities
1 · Linear equations
One variable, power one
ax + b = 0, a ≠ 0
Degree one. The exponent of the variable is always 1. So x2 = 4 is not linear, and neither is xy = 14 (a product of two variables).
Solution. A value of the variable that makes both sides equal.
Equivalent equations have the same solutions. Adding or subtracting the same number on both sides keeps them equivalent.
2x + 3 = 4 ⇔ 2x = 1
2x + 3 = 4 ⇔ 2x = 1
2 · Solving
Clear fractions and decimals first
−7x + 24 = 3
Subtract 24: −7x = −21. Divide by −7: x = 3.
Subtract 24: −7x = −21. Divide by −7: x = 3.
Fractions. 3x/5 − 1/2 = x/4 + 1
Multiply every term by 20, the LCM of the denominators, so every term becomes whole: 12x − 10 = 5x + 20. Then 7x = 30 and x = 30/7.
Multiply every term by 20, the LCM of the denominators, so every term becomes whole: 12x − 10 = 5x + 20. Then 7x = 30 and x = 30/7.
Decimals. 0.7(x − 1) − 0.5x = 1.1
Expand: 0.2x − 0.7 = 1.1, so 0.2x = 1.8 and x = 9.
Expand: 0.2x − 0.7 = 1.1, so 0.2x = 1.8 and x = 9.
3 · Special cases
When there is no solution
2x + 3 = 2x + 5
The variable cancels and leaves 3 = 5, which is false. No value of x works.
The variable cancels and leaves 3 = 5, which is false. No value of x works.
We then write the solution set as φ, the empty set.
4 · Absolute value
Distance from zero
|x| is the distance of x from 0, so it is never negative. |−5| = 5.
Two cases. |x| = x when x ≥ 0, and |x| = −x when x < 0.
Products. |ab| = |a| |b|
Equal absolute values. If |a| = |b|, then a = b or a = −b, since both are the same distance from 0.
|x| = 0 has exactly one solution, x = 0.
5 · Absolute value equations
Solve two equations
|ax + b| = c (c > 0)
The quantity inside the bars can be c or −c, so solve ax + b = c and ax + b = −c.
|2x + 5| = 11
2x + 5 = 11 gives x = 3. 2x + 5 = −11 gives x = −8.
2x + 5 = 11 gives x = 3. 2x + 5 = −11 gives x = −8.
|3x + 2| = 7
3x + 2 = 7 gives x = 5/3. 3x + 2 = −7 gives x = −3.
3x + 2 = 7 gives x = 5/3. 3x + 2 = −7 gives x = −3.
Negative on the right. |x| = −6 has no solution. Likewise |x| + 7 = 3 gives |x| = −4, which is impossible.
6 · Radical equations
Square, then always check
A radical equation has the variable under a radical sign, such as √(x + 1) = 7.
√(2x + 1) = 9
Square both sides: 2x + 1 = 81, so x = 40. Check: √81 = 9 ✓.
Square both sides: 2x + 1 = 81, so x = 40. Check: √81 = 9 ✓.
√x + 3 = 7
Isolate the root: √x = 4, so x = 16. Check: √16 + 3 = 7 ✓.
Isolate the root: √x = 4, so x = 16. Check: √16 + 3 = 7 ✓.
Extraneous roots. Squaring can create a root that fails the original equation. Such a root is called extraneous and is rejected.
No solution. √(3y + 1) = −2 is impossible, because a principal square root is never negative. In 4 + 2√(3y + 1) = 3, squaring gives y = −¼, but that fails the check, so there is no solution.
7 · Inequalities
Solving linear inequalities
A linear inequality in one variable has the form ax + b < 0 (or >, ≤, ≥) with a ≠ 0. Strict inequalities use only < and >.
x + 5 < 8: subtract 5 from both sides, giving x < 3.
3x − 2 < 7 gives 3x < 9, so x < 3. The value 3 is not a solution, because at x = 3 the left side equals 7, which is not less than 7.
8 · The key rule
Negative number? Flip the sign
When you multiply or divide both sides of an inequality by a negative number, the inequality sign reverses.
−2x < 6
Divide by −2 and flip: x > −3.
Divide by −2 and flip: x > −3.
−3x ≥ 12
Divide by −3 and flip: x ≤ −4.
Divide by −3 and flip: x ≤ −4.
The usual slip. Writing x < −3 for the first one. Check with a number: x = 0 satisfies −2(0) < 6, but 0 < −3 is false.
9 · Compound inequalities
“And” and “or”
and
A number must satisfy both. The solution is the common part.
or
One is enough. The solution joins the two sets.
And. x − 5 ≥ −1 and x + 3 ≤ 10 give x ≥ 4 and x ≤ 7, so 4 ≤ x ≤ 7.
Or. 2x + 3 > 7 or 4x − 1 < 3 give x > 2 or x < 1. Join the two sets.
Combined form. 4 < 2x + 3 < 10 means two inequalities joined by “and”. For “twice a number plus 3 is between 5 and 17”: 5 < 2x + 3 < 17, so 1 < x < 7.
10 · Common mistakes
Five traps to avoid
Not flipping the sign after dividing by a negative number.
Giving one answer for |x| = k when k > 0. There are two: k and −k. And if k < 0 there are none.
Skipping the check in a radical equation. Squaring can introduce an extraneous root.
Including the end value of a strict inequality. For 3x − 2 < 7, x = 3 is not a solution.
Mixing up “and” with “or”. “And” keeps only the overlap; “or” keeps everything in either set.
11 · Quick check
Try these, then tap to check
12 · Your turn
Practise Unit 5
Three timed MCQ tests with an explanation after every answer.
Practice questions are independently written and are not official FBISE papers.