\(2y - 3 = 7\)
\begin{align*}
2y - 3 & = 7 \\
2y & = 10 \\
y & = 5
\end{align*}
4.2 Solving linear equations
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4.1 Introduction
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4.3 Solving quadratic equations
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4.2 Solving linear equations (EMA34)
The simplest equation to solve is a linear equation. A linear equation is an equation where the highest exponent of the variable is \(\text{1}\). The following are examples of linear equations:
\begin{align*} 2x + 2 & = 1 \\ \frac{2 - x}{3x + 1} & = 2 \\ 4\left(2x - 9\right) - 4x & = 4 - 6x \\ \frac{2a - 3}{3} - 3a & = \frac{a}{3} \end{align*}Solving an equation means finding the value of the variable that makes the equation true. For example, to solve the simple equation \(x + 1 = 1\), we need to determine the value of \(x\) that will make the left hand side equal to the right hand side. The solution is \(x = 0\).
The solution, also called the root of an equation, is the value of the variable that satisfies the equation. For linear equations, there is at most one solution for the equation.
To solve equations we use algebraic methods that include expanding expressions, grouping terms, and factorising.
For example:
\begin{align*} 2x + 2 & = 1 \\ 2x & =1 - 2 \quad \text{ (rearrange)} \\ 2x & = -1 \quad \text{ (simplify)} \\ x & = -\frac{1}{2} \quad \text{(divide both sides by } 2\text{)} \end{align*}Check the answer by substituting \(x=-\frac{1}{2}\).
\begin{align*} \text{LHS } & = 2x + 2 \\ & = 2\left(-\frac{1}{2}\right) + 2 \\ & = -1 + 2 \\ & = 1 \\ \text{RHS } & =1 \end{align*}Therefore \(x=-\frac{1}{2}\)
The following video gives an introduction to solving linear equations.
Method for solving linear equations (EMA35)
The general steps for solving linear equations are:
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Expand all brackets.
-
Rearrange the terms so that all terms containing the variable are on one side of the equation and all constant terms are on the other side.
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Group like terms together and simplify.
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Factorise if necessary.
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Find the solution and write down the answer.
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Check the answer by substituting the solution back into the original equation.
An equation must always be balanced, whatever you do to the left-hand side, you must also do to the right-hand side.
Worked example 1: Solving linear equations
Solve for \(x\):
\[4(2x - 9) - 4x = 4 - 6x\]Expand the brackets and simplify
\begin{align*} 4(2x - 9) - 4x & = 4 - 6x \\ 8x - 36 - 4x & = 4 - 6x \\ 8x - 4x + 6x & = 4 + 36 \\ 10x & = 40 \end{align*}Divide both sides by 10
\[x = 4\]Check the answer by substituting the solution back into the original equation
\begin{align*} \text{LHS } & = 4[2(4) - 9] - 4(4) \\ & = 4(8 - 9) - 16 \\ & = 4(-1) - 16 \\ & = -4 - 16 \\ & = -20 \\ \text{RHS } & = 4 - 6(4) \\ & =4 - 24 \\ & = -20 \\ \therefore \text{LHS } = \text{ RHS} \end{align*}Since both sides are equal, the answer is correct.
Worked example 2: Solving linear equations
Solve for \(x\):
\[\frac{2 - x}{3x + 1} = 2\]Multiply both sides of the equation by \(\left(3x + 1\right)\)
Division by \(\text{0}\) is undefined so there must be a restriction: \(\left(x\ne -\frac{1}{3}\right)\).
\begin{align*} \frac{2 - x}{3x + 1} & = 2 \\ (2 - x) & = 2(3x + 1) \end{align*}Expand the brackets and simplify
\begin{align*} 2 - x & = 6x + 2 \\ -x - 6x & = 2 - 2 \\ -7x & = 0 \end{align*}Divide both sides by \(-\text{7}\)
\begin{align*} x & = \frac{0}{-7} \\ x & = 0 \end{align*}Check the answer by substituting the solution back into the original equation
\begin{align*} \text{LHS } & = \frac{2 - (0)}{3(0) + 1} \\ & = 2 \\ & = \text{ RHS} \end{align*}Since both sides are equal, the answer is correct.
Worked example 3: Solving linear equations
Solve for \(a\): \[\frac{2a - 3}{3} - 3a = \frac{a}{3}\]
Multiply the equation by the common denominator \(\text{3}\) and simplify
\begin{align*} 2a - 3 - 9a & = a \\ -7a - 3 & = a \end{align*}Rearrange the terms and simplify
\begin{align*} -7a - a & = 3 \\ -8a & = 3 \end{align*}Divide both sides by \(-\text{8}\)
\[a= -\frac{3}{8}\]Check the answer by substituting the solution back into the original equation
\begin{align*} \text{LHS } & = \frac{2\left(-\frac{3}{8}\right) - 3}{3} - 3\left(-\frac{3}{8}\right) \\ & = \frac{\left(-\frac{3}{4}\right) - \frac{12}{4}}{3} + \frac{9}{8} \\ & = \left[-\frac{15}{4}\times \frac{1}{3}\right] + \frac{9}{8} \\ & = -\frac{5}{4} + \frac{9}{8} \\ & = -\frac{10}{8} + \frac{9}{8} \\ & = -\frac{1}{8} \\ \text{RHS } & = \frac{-\frac{3}{8}}{3} \\ & = \frac{-\frac{3}{8}}{3} \\ & = -\frac{3}{8}\times \frac{1}{3} \\ & = -\frac{1}{8} \\ \therefore \text{LHS } = \text{ RHS} \end{align*}Since both sides are equal, the answer is correct.
Textbook Exercise 4.1
Solve the following equations (assume all denominators are non-zero):
\(2c = c -8\)
\begin{align*}
2c &= c - 8\\
c & = -8
\end{align*}
\(3 = 1 - 2c\)
\begin{align*} 3 &= 1 - 2c\\ 2c &= 1 - (3)\\ 2c & = -2\\ c & = \frac{-2}{2}\\ & = -1 \end{align*}
\(4b+5 = -7\)
\begin{align*} 4b +5 &= -7\\ 4b &= -7 - (5)\\ 4b & = -12\\ b & = \frac{-12}{4}\\ & = -3 \end{align*}
\(-3y = 0\)
\begin{align*}
-3y & = 0 \\
y & = 0
\end{align*}
\(16y + 4 = -10\)
\begin{align*}
16y + 4 & = -10 \\
16y & = -14 \\
y & = -\frac{14}{16}\\
& = -\frac{7}{8}
\end{align*}
\(12y + 0 = 144\)
\begin{align*}
12y + 0 & = 144 \\
12y & = 144 \\
y & = 12
\end{align*}
\(7 + 5y = 62\)
\begin{align*}
7 + 5y & = 62 \\
5y & = 55 \\
y & = 11
\end{align*}
\(55 = 5x + \frac{3}{4}\)
\begin{align*}
55 & = 5x + \frac{3}{4} \\
220 & = 20x + 3 \\
20x & = 217 \\
x & = \frac{217}{20}
\end{align*}
\(5x = 2x + 45\)
\begin{align*}
5x & = 2x + 45 \\
3x & = 45 \\
x & = 15
\end{align*}
\(23x - 12 = 6 + 3x\)
\begin{align*}
23x - 12 & = 6 + 3x \\
20x & = 18 \\
x & = \frac{18}{20} \\
& = \frac{9}{10}
\end{align*}
\(12 - 6x + 34x = 2x - 24 - 64\)
\begin{align*}
12 - 6x + 34x & = 2x - 24 - 64 \\
12 + 28x & = 2x - 88 \\
26x & = -100 \\
x & = -\frac{100}{26} \\
& = -\frac{50}{13}
\end{align*}
\(6x + 3x = 4 - 5(2x - 3)\)
\begin{align*}
6x + 3x & = 4 - 5(2x - 3) \\
9x & = 4 - 10x + 15 \\
19x & = 19 \\
x & = 1
\end{align*}
\(18 - 2p = p + 9\)
\begin{align*}
18 - 2p & = p + 9 \\
9 & = 3p \\
p & = 3
\end{align*}
\(\dfrac{4}{p} = \dfrac{16}{24}\)
\begin{align*}
\frac{4}{p} & = \frac{16}{24} \\
(4)(24) & = (16)(p) \\
16p & = 96 \\
p & = 6
\end{align*}
\(-(-16 - p) = 13p - 1\)
\begin{align*}
-(-16 - p) & = 13p - 1 \\
16 + p & = 13p - 1 \\
17 & = 12p \\
p & = \frac{17}{12}
\end{align*}
\(3f - 10 = 10\)
\begin{align*}
3f - 10 & = 10 \\
3f & = 20 \\
f & = \frac{20}{3}
\end{align*}
\(3f + 16 = 4f - 10\)
\begin{align*}
3f + 16 & = 4f - 10 \\
f & = 26
\end{align*}
\(10f + 5 = -2f -3f + 80\)
\begin{align*}
10f + 5 & = -2f - 3f + 80 \\
10f + 5 & = -5f + 80 \\
15f & = 75 \\
f & = 5
\end{align*}
\(8(f - 4) = 5(f - 4)\)
\begin{align*}
8(f - 4) & = 5( f - 4) \\
8f - 32 & = 5f - 20 \\
3f & = 12 \\
f & = 4
\end{align*}
\(6 = 6(f + 7) + 5f\)
\begin{align*}
6 & = 6(f + 7) + 5f \\
6 & = 6f + 42 + 5f \\
-36 & = 11f \\
f & = -\frac{36}{11}
\end{align*}
\(-7x = 8(1 - x)\)
\begin{align*}
-7x & = 8(1 - x) \\
-7x & = 8 - 8x \\
x & = 8
\end{align*}
\(5 - \dfrac{7}{b} = \dfrac{2(b + 4)}{b}\)
\begin{align*}
5 - \frac{7}{b} & = \frac{2(b + 4)}{b} \\
\frac{5b - 7}{b} & = \frac{2b + 8}{b} \\
5b - 7 & = 2b + 8 \\
3b & = 15\\
b & = 5
\end{align*}
\(\dfrac{x + 2}{4} - \dfrac{x - 6}{3} = \dfrac{1}{2}\)
\begin{align*}
\frac{x + 2}{4} - \frac{x - 6}{3} & = \frac{1}{2} \\
\frac{3(x + 2) - 4(x - 6)}{12} & = \frac{1}{2} \\
\frac{3x + 6 - 4x + 24}{12} & = \frac{1}{2} \\
(-x + 30)(2) & = 12 \\
-2x + 60 & = 12\\
-2x & = -48 \\
x & = 24
\end{align*}
\(1 = \dfrac{3a - 4}{2a + 6}\)
Note that \(a \neq - -3\)
\begin{align*} 1 &= \frac{3a - 4}{2a + 6} \\ 2a + 6 &= 3a - 4 \\ a &= 10 \end{align*}
\(\dfrac{2-5a}{3} - 6 = \dfrac{4a}{3} +2 - a\)
\begin{align*}
\frac{2-5a}{3} - 6 &= \frac{4a}{3} +2 - a \\
\frac{2-5a}{3} - \frac{4a}{3} + a &= 8 \\
\frac{2-5a - 4 a + 3a}{3} &= 8 \\
2 - 6a &= 24 \\
6a &= -22 \\
a &= -\frac{22}{6}
\end{align*}
\(2 - \dfrac{4}{b+5} = \dfrac{3b}{b+5}\)
Note \(b \neq -5\)
\begin{align*} 2 - \frac{4}{b+5} &= \frac{3b}{b+5} \\ 2 &= \frac{3b+4}{b+5} \\ 2b + 10 &= 3b + 4 \\ b &= 6 \end{align*}\(3 - \dfrac{y - 2}{4} = 4\)
\begin{align*}
3 - \frac{y - 2}{4} & = 4 \\
-\frac{y - 2}{4} & = 1 \\
-y + 2 & = 4 \\
y & = -2
\end{align*}
\(\text{1,5}x + \text{3,125} = \text{1,25}x\)
\begin{align*}
\text{1,5}x + \text{3,125} & = \text{1,25}x \\
\text{1,5}x - \text{1,25}x & = -\text{3,125} \\
\text{0,25}x & = -\text{3,125} \\
x & = -\text{12,5}
\end{align*}
\(\text{1,3}(\text{2,7}x + 1) = \text{4,1} - x\)
\begin{align*}
\text{1,3}(\text{2,7}x + 1) &= \text{4,1} - x \\
\text{3,51}x + \text{1,3} &= \text{4,1} - x \\
\text{4,51}x &= \text{2,8} \\
x & = \frac{\text{2,8}}{\text{4,51}} \\
&= \frac{280}{451}
\end{align*}
\(\text{6,5}x - \text{4,15}= 7 + \text{4,25}x\)
\begin{align*}
\text{6,5}x - \text{4,15} &= 7 + \text{4,25}x \\
\text{2,25}x &= \text{11,15} \\
x & = \frac{\text{11,15}}{\text{2,25}} \\
& = \frac{\text{1 115}}{225} \\
&= \frac{223}{45}
\end{align*}
\(\frac{1}{3}P + \frac{1}{2}P - 10 = 0\)
\begin{align*}
\frac{1}{3}P + \frac{1}{2}P - 10 & = 0 \\
\frac{2 + 3}{6}P & = 10 \\
5P & = 60 \\
P & = 12
\end{align*}
\(1\frac{1}{4}(x - 1) - 1\frac{1}{2}(3x + 2) = 0\)
\begin{align*}
1\frac{1}{4}(x - 1) - 1\frac{1}{2}(3x + 2) & = 0 \\
\frac{5}{4}x - \frac{5}{4} - \frac{3}{2}(3x) - \frac{3}{2}(2) & = 0 \\
\frac{5}{4}x - \frac{5}{4} - \frac{9}{2}x - \frac{6}{2} & = 0 \\
\frac{5 - 18}{4}x + \frac{-5 - 12}{4} & = 0 \\
\frac{-13}{4}x & = \frac{17}{4} \\
-13x & = 17 \\
x & = -\frac{17}{13}
\end{align*}
\(\frac{1}{5}(x- 1) = \frac{1}{3}(x-2) + 3\)
\begin{align*}
\frac{1}{5}(x- 1) &= \frac{1}{3}(x-2) + 3 \\
\frac{1}{5}x- \frac{1}{5} &= \frac{1}{3}x- \frac{2}{3} + 3 \\
-\frac{1}{5} + \frac{2}{3} - 3 &= \frac{2}{15}x \\
-\frac{38}{15} &= \frac{2}{15}x \\
x &= -\frac{38}{2} \\
x &= -19
\end{align*}
\(\dfrac{5}{2a} + \dfrac{1}{6a} - \dfrac{3}{a} = 2\)
\begin{align*}
\frac{5}{2a} + \frac{1}{6a} - \frac{3}{a} & = 2 \\
\frac{5(3) + 1 - 3(6)}{6a} & = 2 \\
\frac{15 + 1 - 18}{6a} & = 2 \\
\frac{-2}{6a} & = 2 \\
-2 & = 12a \\
a & = -\frac{1}{6}
\end{align*}
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4.3 Solving quadratic equations
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