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question 5: solve the following equations (a) $\frac{3x + 5}{2}=7$ (b) …

Question

question 5: solve the following equations
(a) $\frac{3x + 5}{2}=7$ (b) $\frac{5x - 12}{3}=11$ (c) $\frac{4x + 2}{6}=5$ (d) $\frac{10x + 3}{4}=4$
(e) $\frac{5x - 8}{2}=10$ (f) $\frac{8x + 4}{5}=12.8$ (g) $\frac{2x + 13}{3}=1$ (h) $\frac{3x - 4}{7}=-4$
(i) $\frac{7x - 12}{3}=-25$ (j) $\frac{29 - 2x}{3}=5$ (k) $\frac{100 - 5x}{3}=30$ (l) $\frac{24 - 3x}{12}=5$
question 6: solve the equations below
(a) $\frac{2x + 1}{3}=x - 2$ (b) $\frac{5x - 3}{2}=2x + 4$ (c) $\frac{x + 17}{5}=x + 1$
(d) $4x - 9=\frac{2x + 3}{3}$ (e) $\frac{9 - x}{2}=x + 3$ (f) $\frac{15 - 2x}{3}=2x - 3$
question 7: solve the equations below
(a) $\frac{2x - 1}{x + 3}=9$ (b) $\frac{x + 11}{2x - 5}=2$ (c) $\frac{x + 1}{x + 4}=3$
(d) $\frac{5x - 24}{x - 4}=3$ (e) $\frac{x + 7}{x - 3}=-4$ (f) $\frac{3x + 8}{9 - x}=-2$
question 8: solve the equations below
(a) $\frac{x + 3}{2}+\frac{x + 1}{4}=10$ (b) $\frac{x + 3}{10}+\frac{x - 2}{5}=2$ (c) $\frac{2x - 1}{9}+\frac{x + 2}{3}=0$
(d) $\frac{x - 1}{4}-\frac{x + 3}{2}=-4$ (e) $\frac{4x + 9}{15}-\frac{x - 3}{5}=1$ (f) $\frac{x + 4}{3}+\frac{x + 1}{2}=1$
(g) $\frac{3x + 5}{4}-\frac{x - 7}{5}=1$ (h) $\frac{2x - 5}{7}-\frac{2x - 1}{2}=3$ (i) $\frac{x}{2}+\frac{4x + 1}{10}=-8$
(j) $\frac{x + 1}{2}+\frac{2x - 1}{4}+\frac{x + 2}{3}=1$

Explanation:

Response
  1. Answer - Question 5(a):
  • First, we start with the equation \(\frac{3x + 5}{2}=7\).
  • Multiply both - sides of the equation by 2 to get rid of the denominator: \(3x+5 = 7\times2\).
  • \(3x + 5=14\).
  • Then subtract 5 from both sides: \(3x=14 - 5\).
  • \(3x = 9\).
  • Finally, divide both sides by 3: \(x=\frac{9}{3}=3\).
  1. Answer - Question 5(b):
  • Given \(\frac{5x - 12}{3}=11\).
  • Multiply both sides by 3: \(5x-12 = 11\times3\).
  • \(5x-12 = 33\).
  • Add 12 to both sides: \(5x=33 + 12\).
  • \(5x = 45\).
  • Divide both sides by 5: \(x=\frac{45}{5}=9\).
  1. Answer - Question 5(c):
  • For \(\frac{4x + 2}{6}=5\).
  • Multiply both sides by 6: \(4x+2 = 5\times6\).
  • \(4x+2 = 30\).
  • Subtract 2 from both sides: \(4x=30 - 2\).
  • \(4x = 28\).
  • Divide both sides by 4: \(x=\frac{28}{4}=7\).
  1. Answer - Question 5(d):
  • Given \(\frac{10x + 3}{4}=4\).
  • Multiply both sides by 4: \(10x+3 = 4\times4\).
  • \(10x+3 = 16\).
  • Subtract 3 from both sides: \(10x=16 - 3\).
  • \(10x = 13\).
  • Divide both sides by 10: \(x=\frac{13}{10}=1.3\).
  1. Answer - Question 5(e):
  • For \(\frac{5x - 8}{2}=10\).
  • Multiply both sides by 2: \(5x-8 = 10\times2\).
  • \(5x-8 = 20\).
  • Add 8 to both sides: \(5x=20 + 8\).
  • \(5x = 28\).
  • Divide both sides by 5: \(x=\frac{28}{5}=5.6\).
  1. Answer - Question 5(f):
  • Given \(\frac{8x + 4}{5}=12.8\).
  • Multiply both sides by 5: \(8x+4 = 12.8\times5\).
  • \(8x+4 = 64\).
  • Subtract 4 from both sides: \(8x=64 - 4\).
  • \(8x = 60\).
  • Divide both sides by 8: \(x=\frac{60}{8}=7.5\).
  1. Answer - Question 5(g):
  • For \(\frac{2x + 13}{3}=1\).
  • Multiply both sides by 3: \(2x+13 = 1\times3\).
  • \(2x+13 = 3\).
  • Subtract 13 from both sides: \(2x=3 - 13\).
  • \(2x=-10\).
  • Divide both sides by 2: \(x=\frac{-10}{2}=-5\).
  1. Answer - Question 5(h):
  • Given \(\frac{3x - 4}{7}=-4\).
  • Multiply both sides by 7: \(3x-4=-4\times7\).
  • \(3x-4=-28\).
  • Add 4 to both sides: \(3x=-28 + 4\).
  • \(3x=-24\).
  • Divide both sides by 3: \(x=\frac{-24}{3}=-8\).
  1. Answer - Question 5(i):
  • For \(\frac{7x - 12}{3}=-25\).
  • Multiply both sides by 3: \(7x-12=-25\times3\).
  • \(7x-12=-75\).
  • Add 12 to both sides: \(7x=-75 + 12\).
  • \(7x=-63\).
  • Divide both sides by 7: \(x=\frac{-63}{7}=-9\).
  1. Answer - Question 5(j):
  • Given \(\frac{29 - 2x}{3}=5\).
  • Multiply both sides by 3: \(29-2x = 5\times3\).
  • \(29-2x = 15\).
  • Subtract 29 from both sides: \(-2x=15 - 29\).
  • \(-2x=-14\).
  • Divide both sides by - 2: \(x=\frac{-14}{-2}=7\).
  1. Answer - Question 5(k):
  • For \(\frac{100 - 5x}{3}=30\).
  • Multiply both sides by 3: \(100-5x = 30\times3\).
  • \(100-5x = 90\).
  • Subtract 100 from both sides: \(-5x=90 - 100\).
  • \(-5x=-10\).
  • Divide both sides by - 5: \(x=\frac{-10}{-5}=2\).
  1. Answer - Question 5(l):
  • Given \(\frac{24 - 3x}{12}=5\).
  • Multiply both sides by 12: \(24-3x = 5\times12\).
  • \(24-3x = 60\).
  • Subtract 24 from both sides: \(-3x=60 - 24\).
  • \(-3x = 36\).
  • Divide both sides by - 3: \(x=\frac{36}{-3}=-12\).

Since there are too many sub - questions in this problem to list all solutions for all questions, the general method for solving linear equations of the form \(\frac{ax + b}{c}=d\) (as in Question 5) is:

Step1: Eliminate denominator

Multiply both sides by the denominator value. For \(\frac{ax + b}{c}=d\), we get \(ax + b=cd\).

Step2: Isolate the term with \(x\)

Use addition or subtraction to move the constant term to the other side. For \(ax + b=cd\), we get \(ax=cd - b\).

Step3: Solve for \(x\)

Divide both sides by the coefficient of \(x\). So \(x=\frac{cd - b}{a}\).

Answer:

(a) \(x = 3\); (b) \(x = 9\); (c) \(x = 7\); (d) \(x = 1.3\); (e) \(x = 5.6\); (f) \(x = 7.5\); (g) \(x=-5\); (h) \(x=-8\); (i) \(x=-9\); (j) \(x = 7\); (k) \(x = 2\); (l) \(x=-12\)