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Linear Equations

Chapter 5 Study Notes · O-Level 4052

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Section 2 of 7
Contents▾
Why Equations Matter5.1Linear Equations: Concept and Solving5.2Linear Equations with Fractional Coefficients5.3Fractional Equations5.4Applications: Word Problems5.5Mathematical FormulaeChapter Summary

5.1Linear Equations: Concept and Solving

Recall: Chapter 4 (Basic Algebra), algebraic expressions, expansion, and like terms.→

An equation says two expressions are equal. To solve it is to find the value of the unknown that makes that true.

The balancing principle

Do the same thing to both sides. An equation is a balance: change one side alone and it stops balancing.

Balancing principle
a=b  ⇒  a±c=b±c,ac=bc,ac=bc  (c≠0)a = b \;\Rightarrow\; a \pm c = b \pm c, \quad ac = bc, \quad \frac{a}{c} = \frac{b}{c} \; (c \neq 0)
  • Add or subtract the same number on both sides to move a constant.
  • Multiply or divide both sides by the same non-zero number to move a coefficient.
  • Expand any brackets first, then collect the unknown on one side.

Two-step equation

Solve 5x−7=18\displaystyle 5x - 7 = 18.

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5x−7\displaystyle 5x - 7
=18\displaystyle = 18
5x\displaystyle 5x
=25add 7 to both sides\displaystyle = 25 \quad \scriptsize\textit{add 7 to both sides}
x\displaystyle x
=5divide both sides by 5\displaystyle = 5 \quad \scriptsize\textit{divide both sides by 5}

Now you try

Solve y+11=4y + 11 = 4.

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y+11\displaystyle y + 11
=4\displaystyle = 4
y\displaystyle y
=4−11subtract 11 from both sides\displaystyle = 4 - 11 \quad \scriptsize\textit{subtract 11 from both sides}
y\displaystyle y
=−7\displaystyle = -7

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Variables on both sides

Solve 7y−3=4y+15\displaystyle 7y - 3 = 4y + 15.

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7y−3\displaystyle 7y - 3
=4y+15\displaystyle = 4y + 15
3y−3\displaystyle 3y - 3
=15subtract 4y from both sides\displaystyle = 15 \quad \scriptsize\textit{subtract 4y from both sides}
3y\displaystyle 3y
=18add 3 to both sides\displaystyle = 18 \quad \scriptsize\textit{add 3 to both sides}
y\displaystyle y
=6\displaystyle = 6

Equation with brackets

Solve 3(2m+1)=4(m−5)\displaystyle 3(2m + 1) = 4(m - 5).

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6m+3\displaystyle 6m + 3
=4m−20expand both sides\displaystyle = 4m - 20 \quad \scriptsize\textit{expand both sides}
2m\displaystyle 2m
=−23collect terms\displaystyle = -23 \quad \scriptsize\textit{collect terms}
m\displaystyle m
=−232\displaystyle = -\frac{23}{2}

Now you try

Solve 2(x+5)=62(x + 5) = 6.

Show solution▾
2(x+5)\displaystyle 2(x + 5)
=6\displaystyle = 6
2x+10\displaystyle 2x + 10
=6expand\displaystyle = 6 \quad \scriptsize\textit{expand}
2x\displaystyle 2x
=6−10subtract 10 from both sides\displaystyle = 6 - 10 \quad \scriptsize\textit{subtract 10 from both sides}
2x\displaystyle 2x
=−4\displaystyle = -4
x\displaystyle x
=−42\displaystyle = -\frac{4}{2}
x\displaystyle x
=−2\displaystyle = -2

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Reverse problem: find the coefficient

The solution to ax−3=2x+9\displaystyle ax - 3 = 2x + 9 is x=4\displaystyle x = 4. Find a\displaystyle a.

Show solution▾
Substitute x=4x = 4 into the equation:
a(4)−3\displaystyle a(4) - 3
=2(4)+9\displaystyle = 2(4) + 9
4a−3\displaystyle 4a - 3
=8+9\displaystyle = 8 + 9
4a−3\displaystyle 4a - 3
=17\displaystyle = 17
4a\displaystyle 4a
=20\displaystyle = 20
a\displaystyle a
=5\displaystyle = 5
Check: 5(4)−3=175(4) - 3 = 17 and 2(4)+9=172(4) + 9 = 17 ✓

Practise this

Questions on Linear Equations, marked as you go, with the working shown.

Practise Linear Equations