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Mathematics

Solve the following linear equations:

(i) 5x − 6 = 12 − x

(ii) n3+1=4−n\dfrac{n}{3} + 1 = 4 - n

(iii) 5p + 7 = 19 − 2p

(iv) 2x+23=52−x2x + \dfrac{2}{3} = \dfrac{5}{2} - x

(v) x2−5=x3−4\dfrac{x}{2} - 5 = \dfrac{x}{3} - 4

(vi) 18−3y4=11+y18 - \dfrac{3y}{4} = 11 + y

Simple Equations

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Answer

(i) We have:

5x − 6 = 12 − x

⇒ 5x + x = 12 + 6     [Transposing −x to LHS and −6 to RHS]

⇒ 6x = 18

⇒ x=186x = \dfrac{18}{6}

∴ x = 3.

(ii) We have:

⇒n3+1=4−n⇒n3+n=4−1⇒n+3n3=3⇒4n3=3⇒4n=9⇒n=94=214\Rightarrow \dfrac{n}{3} + 1 = 4 - n \\[1em] \Rightarrow \dfrac{n}{3} + n = 4 - 1 \\[1em] \Rightarrow \dfrac{n + 3n}{3} = 3 \\[1em] \Rightarrow \dfrac{4n}{3} = 3 \\[1em] \Rightarrow 4n = 9 \\[1em] \Rightarrow n = \dfrac{9}{4} = 2\dfrac{1}{4}

∴ n = 2142\dfrac{1}{4}

(iii) We have:

5p + 7 = 19 − 2p

⇒ 5p + 2p = 19 − 7     [Transposing −2p to LHS and +7 to RHS]

⇒ 7p = 12

⇒ p=127=157p = \dfrac{12}{7} = 1\dfrac{5}{7}

∴ p = 1571\dfrac{5}{7}.

(iv) We have:

⇒2x+23=52−x⇒2x+x=52−23⇒3x=15−46⇒3x=116⇒x=1118\Rightarrow 2x + \dfrac{2}{3} = \dfrac{5}{2} - x\\[1em] \Rightarrow 2x + x = \dfrac{5}{2} - \dfrac{2}{3}\\[1em] \Rightarrow 3x = \dfrac{15 - 4}{6} \\[1em] \Rightarrow 3x = \dfrac{11}{6} \\[1em] \Rightarrow x = \dfrac{11}{18}

∴ x = 1118\dfrac{11}{18}.

(v) We have:

⇒x2−5=x3−4⇒x2−x3=−4+5⇒3x−2x6=1⇒x6=1⇒x=6\Rightarrow \dfrac{x}{2} - 5 = \dfrac{x}{3} - 4 \\[1em] \Rightarrow \dfrac{x}{2} - \dfrac{x}{3} = -4 + 5 \\[1em] \Rightarrow \dfrac{3x - 2x}{6} = 1 \\[1em] \Rightarrow \dfrac{x}{6} = 1 \\[1em] \Rightarrow x = 6

∴ x = 6.

(vi) We have:

⇒18−3y4=11+y⇒18−11=y+3y4⇒7=4y+3y4⇒7=7y4⇒7y=28⇒y=287=4\Rightarrow 18 - \dfrac{3y}{4} = 11 + y\\[1em] \Rightarrow 18 - 11 = y + \dfrac{3y}{4} \\[1em] \Rightarrow 7 = \dfrac{4y + 3y}{4} \\[1em] \Rightarrow 7 = \dfrac{7y}{4} \\[1em] \Rightarrow 7y = 28 \\[1em] \Rightarrow y = \dfrac{28}{7} = 4

∴ y = 4.

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