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Linear Equations

Chapter 5 Study Notes · O-Level 4052

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

5.4Applications: Word Problems

The skill is turning English into algebra:

Five-step strategy

  1. Define the unknown: "Let xx = ..."
  2. Express other quantities in terms of xx.
  3. Form an equation from the given information.
  4. Solve the equation.
  5. Check and interpret the answer in context.

Movie ticket problem

An adult ticket costs $3\displaystyle \$3 more than a child ticket. Mr Tan buys 2 adult and 3 child tickets for $51\displaystyle \$51. Find the cost of a child ticket.

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Let the child ticket cost $x\$x. Then the adult ticket costs $(x+3)\$(x + 3).
2(x+3)+3x\displaystyle 2(x + 3) + 3x
=51\displaystyle = 51
2x+6+3x\displaystyle 2x + 6 + 3x
=51expand\displaystyle = 51 \quad \scriptsize\textit{expand}
5x\displaystyle 5x
=45\displaystyle = 45
x\displaystyle x
=9\displaystyle = 9
A child ticket costs $9\$9 (and an adult ticket costs $12\$12).

Now you try

The sum of three consecutive even numbers is 3636. Find the largest number.

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Let the smallest even number be nn. define variable\scriptsize\textit{define variable}
The three consecutive even numbers are nn, n+2n + 2 and n+4n + 4.
n+(n+2)+(n+4)\displaystyle n + (n + 2) + (n + 4)
=36\displaystyle = 36
3n+6\displaystyle 3n + 6
=36\displaystyle = 36
3n\displaystyle 3n
=30\displaystyle = 30
n\displaystyle n
=10\displaystyle = 10
Largest number\displaystyle \text{Largest number}
=n+4\displaystyle = n + 4
=10+4\displaystyle = 10 + 4
=14\displaystyle = 14

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Perimeter problem

The length of a rectangle is 5\displaystyle 5 cm more than twice its width. The perimeter is 64\displaystyle 64 cm. Find the dimensions of the rectangle.

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Let the width be ww cm. Then the length is (2w+5)(2w + 5) cm.
Perimeter: 2[w+(2w+5)]=642[w + (2w + 5)] = 64
2(3w+5)\displaystyle 2(3w + 5)
=64simplify inside brackets\displaystyle = 64 \quad \scriptsize\textit{simplify inside brackets}
6w+10\displaystyle 6w + 10
=64\displaystyle = 64
6w\displaystyle 6w
=54\displaystyle = 54
w\displaystyle w
=9\displaystyle = 9
Width =9= 9 cm
Length: 2(9)+5=232(9) + 5 = 23 cm

Now you try

Priya has 44 more 10-cent coins than 20-cent coins. The total value of all the coins is $2.20\$2.20. Find the number of 20-cent coins.

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Let xx be the number of 20-cent coins. define variable\scriptsize\textit{define variable}
Number of 10-cent coins =x+4= x + 4.
Total value in cents: 20x+10(x+4)=22020x + 10(x + 4) = 220
20x+10x+40\displaystyle 20x + 10x + 40
=220\displaystyle = 220
30x+40\displaystyle 30x + 40
=220\displaystyle = 220
30x\displaystyle 30x
=180\displaystyle = 180
x\displaystyle x
=6\displaystyle = 6
The number of 20-cent coins is 66.

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Reverse problem: construct your own equation

Construct a linear equation in x\displaystyle x whose solution is x=−32\displaystyle x = -\dfrac{3}{2}, and verify your equation.

Show solution▾
Start from the solution x=−32x = -\dfrac{3}{2}.
Multiply both sides by 22: 2x=−3\quad 2x = -3.
Add 55 to both sides: 2x+5=2\quad 2x + 5 = 2.
The equation 2x+5=22x + 5 = 2 has solution x=−32x = -\dfrac{3}{2}.
Verify: 2 ⁣(−32)+5=−3+52\!\left(-\dfrac{3}{2}\right) + 5 = -3 + 5
=2= 2 ✓
Note: any reversible step applied to both sides of x=−32x = -\dfrac{3}{2} gives another equation with the same solution. Add or subtract anything; multiply or divide by anything except zero.

Practise this

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

Practise Linear Equations

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