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3 Solving an equation means determining the value of the variable that makes the equation true. We find the values of the variable that make both sides of the equation equal. A linear equation in one variable has one solution. A system of linear equations with two variables requires at least two equations to find both values.

4 <h2>What Is Solving Equations?</h2>

5 What Is Algebra? 🧮 | Simple Explanation with 🎯 Cool Examples for Kids | ✨BrightCHAMPS Math

6 ▶

7 Solving<a>equations</a>is the process<a>of</a>finding the unknown<a>variable</a>that makes both sides of the<a>equation</a>equal. An equation is a mathematical statement where two<a></a><a>expressions</a>, involving a variable, are equal.

8 In these equations, the LHS and RHS can be interchanged, as both sides represent the same value.

9 There are different ways to solve an equation depending on its type, such as linear,<a>quadratic</a>, rational, or radical equations.

10 <h2>How to Solve an Equation?</h2>

11 Solving an equation includes using mathematical operations to isolate the variable to find the value of the unknown variable. This value is found by isolating the variable using mathematical operations. Let’s look at the steps to solve an equation.

12 <ul><li>Addition property of equality: If we<a>add</a>the same<a>number</a>to both sides, the equality is maintained.If $a = b$, then $a + c = b + c$

13 </li>

14 <li>Subtraction property of equality: If we<a>subtract</a>the same number from both sides, the equation remains balanced.If $a = b$, then $a - c = b - c$

15 </li>

16 <li>Multiplication property of equality: If we<a>multiply</a>both sides by the same number, the equality is not affected.If $a = b$, then $ac = bc$, for any number c

17 </li>

18 <li>Division property of equality: If we<a>divide</a>both sides by the same number (except zero), the equality is maintained.If $a = b$, then $\frac ac = \frac bc$ (where $c ≠ 0$).

19 </li>

20 </ul>We isolate the variable on one side of the equation after completing these steps.

21 <h2>How to Solve an Equation with One Variable?</h2>

22 The<a></a><a>linear equation</a>in one variable is expressed in the form $ax + b = 0$, where a and b are<a>real numbers</a>. To solve such equations, follow these steps:

23 <ul><li>Remove<a>parentheses</a>by applying the<a>distributive property</a>, if needed. </li>

24 <li>To simplify the equation, we<a>combine like terms</a>. </li>

25 <li>We eliminate fractional terms from equations by multiplying both sides by the<a></a><a>least common denominator</a>(LCD). </li>

26 <li>If the equation has<a>decimals</a>, multiply both sides by the appropriate<a>power</a>of 10 to convert them into whole numbers. </li>

27 <li>Apply the addition or subtraction property of equality to bring variable terms to one side and constants to the other. </li>

28 <li>We use the multiplication or<a>division property</a>of equality to make the coefficient of the variable equal to 1. </li>

29 <li>Isolate the variable to find the solution. </li>

30 </ul>For example:

31 Solve the equation: $3(x + 4) = 24 + x$

32 Apply the distributive property on the LHS:

33 $→ 3x + 12 = 24 + x$

34 Group the like terms to one side :

35 $→ 3x - x = 24 -12$

36 Simplify both sides:

37 $→ 2x = 12$

38 To isolate x, we divide both sides by 2:

39 $→ x = 6$

40 Solution: $x = 6$

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43 <h2>How to Solve an Equation with the Trial and Error Method?</h2>

42 <h2>How to Solve an Equation with the Trial and Error Method?</h2>

44 Using the trial-and-error method, we test different values of the variable until we find the one that satisfies the equation.

43 Using the trial-and-error method, we test different values of the variable until we find the one that satisfies the equation.

45 For example:

44 For example:

46 Consider the equation $5x = 35$.

45 Consider the equation $5x = 35$.

47 Look for a number that, multiplied by 5, gives 35

46 Look for a number that, multiplied by 5, gives 35

48 We determine $x = 7$ since $5 × 7 = 35$.

47 We determine $x = 7$ since $5 × 7 = 35$.

49 This method works well for simple equations, but for more complicated ones, it can become challenging and time-consuming.

48 This method works well for simple equations, but for more complicated ones, it can become challenging and time-consuming.

50 <h2>How to Solve a Quadratic Equation?</h2>

49 <h2>How to Solve a Quadratic Equation?</h2>

51 Some equations can have more than one solution. This is often the case with<a>quadratic equations</a>, which are equations of degree two. The zeroes of a<a></a><a>quadratic polynomial</a>are the values that satisfy the equation.

50 Some equations can have more than one solution. This is often the case with<a>quadratic equations</a>, which are equations of degree two. The zeroes of a<a></a><a>quadratic polynomial</a>are the values that satisfy the equation.

52 Example:

51 Example:

53 $(x + 3)(x + 2) = 0$

52 $(x + 3)(x + 2) = 0$

54 This is a quadratic equation that can be solved by writing each<a>factor</a>equal to zero:

53 This is a quadratic equation that can be solved by writing each<a>factor</a>equal to zero:

55 $x + 3 = 0 ⟹ x = -3$

54 $x + 3 = 0 ⟹ x = -3$

56 $x + 2 = 0 ⟹ x = -2$

55 $x + 2 = 0 ⟹ x = -2$

57 So, the solutions are $x = -3$ and $x = -2$.

56 So, the solutions are $x = -3$ and $x = -2$.

58 A quadratic equation is generally written in the form:

57 A quadratic equation is generally written in the form:

59 $ax² + bx + c = 0$

58 $ax² + bx + c = 0$

60 When a quadratic equation is solved, up to two roots are obtained: α and β.

59 When a quadratic equation is solved, up to two roots are obtained: α and β.

61 We can solve a quadratic equation in different steps:

60 We can solve a quadratic equation in different steps:

62 <ul><li>Using the<a>completing the square</a>method </li>

61 <ul><li>Using the<a>completing the square</a>method </li>

63 <li>Using the factorization method </li>

62 <li>Using the factorization method </li>

64 <li>Using the<a>formula</a>method</li>

63 <li>Using the<a>formula</a>method</li>

65 </ul><h3>Using Completing the Square Method</h3>

64 </ul><h3>Using Completing the Square Method</h3>

66 Completing the<a>square</a>method systematically solves a quadratic equation by applying the<a></a><a>algebraic identity</a>:

65 Completing the<a>square</a>method systematically solves a quadratic equation by applying the<a></a><a>algebraic identity</a>:

67 $(a + b)^2 = a^2 + 2ab + b^2$

66 $(a + b)^2 = a^2 + 2ab + b^2$

68 <ul><li>We first need to express the equation in<a>standard form</a>:$ax^2 + bx + c = 0.$

67 <ul><li>We first need to express the equation in<a>standard form</a>:$ax^2 + bx + c = 0.$

69 </li>

68 </li>

70 <li>Divide the entire equation by ‘a’. </li>

69 <li>Divide the entire equation by ‘a’. </li>

71 <li>Shift the<a>constant</a><a>term</a>to one side of the equation. </li>

70 <li>Shift the<a>constant</a><a>term</a>to one side of the equation. </li>

72 <li>Add the square of half the<a>coefficient</a>of x to both sides. </li>

71 <li>Add the square of half the<a>coefficient</a>of x to both sides. </li>

73 <li>Complete the left-hand side as a<a>perfect square</a>. </li>

72 <li>Complete the left-hand side as a<a>perfect square</a>. </li>

74 <li>Take the square root of both sides. </li>

73 <li>Take the square root of both sides. </li>

75 <li>Determine the value of x.</li>

74 <li>Determine the value of x.</li>

76 </ul><h3>Using the Factorization Method</h3>

75 </ul><h3>Using the Factorization Method</h3>

77 A quadratic equation can be solved using the<a>factorization</a>method as discussed below:

76 A quadratic equation can be solved using the<a>factorization</a>method as discussed below:

78 <ul><li>First, express the equation in standard form: $ax² + bx + c = 0$ </li>

77 <ul><li>First, express the equation in standard form: $ax² + bx + c = 0$ </li>

79 <li>Split the middle term:Break the middle term, bx, into two terms such that:

78 <li>Split the middle term:Break the middle term, bx, into two terms such that:

80 Their<a>sum</a>equals b

79 Their<a>sum</a>equals b

81 Their<a>product</a>equals $a × c$

80 Their<a>product</a>equals $a × c$

82 </li>

81 </li>

83 </ul>For example:

82 </ul>For example:

84 Solve: $2x² + 19x + 30 = 0$

83 Solve: $2x² + 19x + 30 = 0$

85 Find two numbers that add up to 19 and multiply to 60 (2 × 30)

84 Find two numbers that add up to 19 and multiply to 60 (2 × 30)

86 → 4 and 15

85 → 4 and 15

87 Now rewrite the equation:

86 Now rewrite the equation:

88 $2x² + 4x + 15x + 30 = 0$

87 $2x² + 4x + 15x + 30 = 0$

89 Group and factor:

88 Group and factor:

90 $2x(x + 2) + 15(x + 2) = 0$

89 $2x(x + 2) + 15(x + 2) = 0$

91 Take the<a>common factor</a>:

90 Take the<a>common factor</a>:

92 $(x + 2)(2x + 15) = 0$

91 $(x + 2)(2x + 15) = 0$

93 Now solve each factor:

92 Now solve each factor:

94 $x + 2 = 0 ⇒ x = -2$

93 $x + 2 = 0 ⇒ x = -2$

95 $2x + 15 = 0 ⇒ x = \frac {-15}{2}$

94 $2x + 15 = 0 ⇒ x = \frac {-15}{2}$

96 <h3>Using the Formula Method</h3>

95 <h3>Using the Formula Method</h3>

97 When the equation is of the form $ax² + bx + c = 0$, we use the quadratic formula:

96 When the equation is of the form $ax² + bx + c = 0$, we use the quadratic formula:

98 $x = {-b \pm \sqrt{b^2-4ac} \over 2a}$

97 $x = {-b \pm \sqrt{b^2-4ac} \over 2a}$

99 To find the solution, we substitute the values of a, b, and c into the formula.

98 To find the solution, we substitute the values of a, b, and c into the formula.

100 For example:

99 For example:

101 Solve: $9x² - 12x + 4 = 0$

100 Solve: $9x² - 12x + 4 = 0$

102 Here, $a = 9, b = -12, c = 4$

101 Here, $a = 9, b = -12, c = 4$

103 Apply the quadratic formula:

102 Apply the quadratic formula:

104 $x = {-b \pm \sqrt{b^2-4ac} \over 2a}$

103 $x = {-b \pm \sqrt{b^2-4ac} \over 2a}$

105 $x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4 \times 9 \times 4}}{2 \times 9} $

104 $x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4 \times 9 \times 4}}{2 \times 9} $

106 $x = \frac{12 \pm \sqrt{144 - 144}}{18} $

105 $x = \frac{12 \pm \sqrt{144 - 144}}{18} $

107 $x = \frac{12 \pm \sqrt{0}}{18} $

106 $x = \frac{12 \pm \sqrt{0}}{18} $

108 $x = \frac{12}{18} $

107 $x = \frac{12}{18} $

109 $x = \frac{2}{3} $

108 $x = \frac{2}{3} $

110 Solution: $x = \frac{2}{3} $

109 Solution: $x = \frac{2}{3} $

111 <h2>How to Solve a Rational Equation?</h2>

110 <h2>How to Solve a Rational Equation?</h2>

112 A<a>rational equation</a>has at least one variable in the<a>denominator</a>. To solve it:

111 A<a>rational equation</a>has at least one variable in the<a>denominator</a>. To solve it:

113 <ul><li>We determine a<a>common denominator</a>or<a>cross-multiply</a>. </li>

112 <ul><li>We determine a<a>common denominator</a>or<a>cross-multiply</a>. </li>

114 <li>Solve the resulting equation. </li>

113 <li>Solve the resulting equation. </li>

115 </ul>For example:

114 </ul>For example:

116 Solve: $\frac{2x}{x + 4} = \frac{4}{5} $

115 Solve: $\frac{2x}{x + 4} = \frac{4}{5} $

117 Cross-multiplying gives:

116 Cross-multiplying gives:

118 $5 × 2x = 4(x + 3)$

117 $5 × 2x = 4(x + 3)$

119 $10x = 4x + 12$

118 $10x = 4x + 12$

120 $10x - 4x = 12$

119 $10x - 4x = 12$

121 $6x = 12$

120 $6x = 12$

122 $x = 2$

121 $x = 2$

123 <h2>How to Solve a Radical Equation?</h2>

122 <h2>How to Solve a Radical Equation?</h2>

124 A radical equation is an equation in which the variable is enclosed in a root. To solve it:

123 A radical equation is an equation in which the variable is enclosed in a root. To solve it:

125 <ul><li>Isolate the<a>radical expression</a>. </li>

124 <ul><li>Isolate the<a>radical expression</a>. </li>

126 <li>Remove the radical by squaring both sides. </li>

125 <li>Remove the radical by squaring both sides. </li>

127 <li>Solve the equation obtained.</li>

126 <li>Solve the equation obtained.</li>

128 </ul>Example:

127 </ul>Example:

129 Solve: $\sqrt{(2x - 3)} = 5$

128 Solve: $\sqrt{(2x - 3)} = 5$

130 Square both sides:

129 Square both sides:

131 $(\sqrt{(2x - 3)})² = 5²$

130 $(\sqrt{(2x - 3)})² = 5²$

132 $2x - 3 = 25$

131 $2x - 3 = 25$

133 $2x = 28$

132 $2x = 28$

134 $x = 14$

133 $x = 14$

135 <h2>Tips and Tricks to Master Solving Equations</h2>

134 <h2>Tips and Tricks to Master Solving Equations</h2>

136 Here are some of the tips and tricks that will be helpful for the learners to master solving equations:

135 Here are some of the tips and tricks that will be helpful for the learners to master solving equations:

137 <ol><li>Before solving the equations, read and understand the equations carefully. Identify the unknown and identify the constants and coefficients. </li>

136 <ol><li>Before solving the equations, read and understand the equations carefully. Identify the unknown and identify the constants and coefficients. </li>

138 <li>Always remember to keep the equation balanced. Whatever we do on one side, must be done on the other side too. Add or subtract both sides using the same number. Multiply or divide by the same number on both the sides. </li>

137 <li>Always remember to keep the equation balanced. Whatever we do on one side, must be done on the other side too. Add or subtract both sides using the same number. Multiply or divide by the same number on both the sides. </li>

139 <li>Always try to perform simplification step by step. Combine like terms and remove brackets using the distributive property. Simplify the<a>fractions</a>if necessary. </li>

138 <li>Always try to perform simplification step by step. Combine like terms and remove brackets using the distributive property. Simplify the<a>fractions</a>if necessary. </li>

140 <li>Use inverse operations carefully. Do the opposite operation to isolate the unknown. Addition and<a>subtraction</a>;<a>multiplication</a>and<a>division</a>; powers and roots. </li>

139 <li>Use inverse operations carefully. Do the opposite operation to isolate the unknown. Addition and<a>subtraction</a>;<a>multiplication</a>and<a>division</a>; powers and roots. </li>

141 <li>Always remember to check your solutions. Substitute your answer back into the original equation to ensure that it works. </li>

140 <li>Always remember to check your solutions. Substitute your answer back into the original equation to ensure that it works. </li>

142 </ol><h2>Common Mistakes and How to Avoid Them in Solving Equations</h2>

141 </ol><h2>Common Mistakes and How to Avoid Them in Solving Equations</h2>

143 Solving equations is an important mathematical skill that helps students to develop strong problem-solving skills. However, it is common for students to make small errors that lead to incorrect results. Here are a few common mistakes along with tips to avoid them:

142 Solving equations is an important mathematical skill that helps students to develop strong problem-solving skills. However, it is common for students to make small errors that lead to incorrect results. Here are a few common mistakes along with tips to avoid them:

144 <h2>Real-Life Applications of Solving Equations</h2>

143 <h2>Real-Life Applications of Solving Equations</h2>

145 Solving equations is a fundamental concept in mathematics, and we use it in different fields. Let’s now learn about their importance in real life. Here are a few real-life applications of solving equations.

144 Solving equations is a fundamental concept in mathematics, and we use it in different fields. Let’s now learn about their importance in real life. Here are a few real-life applications of solving equations.

146 <ul><li>Shopping<a>discounts</a>:We can use equations to find the final price of an item after a discount. For example, if an item costs $2000 and is offered at a 50% discount, the final price can be calculated using the equation:Final price = original price - (discount price × original price) x = 2000 - 0.50 × 2000 x = 2000 - 1000 = 1000 So, the final price is $1000.

145 <ul><li>Shopping<a>discounts</a>:We can use equations to find the final price of an item after a discount. For example, if an item costs $2000 and is offered at a 50% discount, the final price can be calculated using the equation:Final price = original price - (discount price × original price) x = 2000 - 0.50 × 2000 x = 2000 - 1000 = 1000 So, the final price is $1000.

147 </li>

146 </li>

148 <li>Business: The<a>profit</a>earned can be calculated using an equation. For example, if earnings = $2,00,000 and expenses = $60,000, the profit can be determined by solving:x = 200000 - 60000, which is $140000 (profit).

147 <li>Business: The<a>profit</a>earned can be calculated using an equation. For example, if earnings = $2,00,000 and expenses = $60,000, the profit can be determined by solving:x = 200000 - 60000, which is $140000 (profit).

149 </li>

148 </li>

150 <li>Travel calculations:This concept can also be used to estimate the time needed to complete a journey. For example: if the total distance is 120 km and the vehicle travels at a speed of 60 km/h, the time can be calculated using the equation:Time = Distance/ Speed = 120/60 = 2 hours (estimated time).

149 <li>Travel calculations:This concept can also be used to estimate the time needed to complete a journey. For example: if the total distance is 120 km and the vehicle travels at a speed of 60 km/h, the time can be calculated using the equation:Time = Distance/ Speed = 120/60 = 2 hours (estimated time).

151 </li>

150 </li>

152 <li>Cooking and recipes: We can use equations to adjust recipes proportionally. These equations can help scale ingredients accurately. </li>

151 <li>Cooking and recipes: We can use equations to adjust recipes proportionally. These equations can help scale ingredients accurately. </li>

153 <li>Daily life problem-solving: These equations can be helpful in sharing food equally whenever necessary. They help a lot in planning different schedules and calculating travel expenses. These often require forming and solving equations.</li>

152 <li>Daily life problem-solving: These equations can be helpful in sharing food equally whenever necessary. They help a lot in planning different schedules and calculating travel expenses. These often require forming and solving equations.</li>

154 - </ul><h3>Problem 1</h3>

153 + </ul><h2>Download Worksheets</h2>

154 + <h3>Problem 1</h3>

155 Solve: 3x + 5 = 20

156 Okay, lets begin

157 $ x = 5$

158 <h3>Explanation</h3>

159 We first subtract 5 from both sides:

160 $3x = 15$

161 Now, divide both sides by 3:

162 $ x = 5$

163 Well explained 👍

164 <h3>Problem 2</h3>

165 Solve: x/3 + 1/2 = 1

166 Okay, lets begin

167 $x = \frac {3}{2}$

168 <h3>Explanation</h3>

169 For equations with fractions, we multiply the whole equation by 6 (LCM of 3 and 2):

170 $6 × (\frac {x}{3} + \frac{1}{2}) = 6× 1$

171 $ (\frac {6 ×x}{3} + \frac{6 ×1}{2}) = 6× 1$

172 $2x + 3 = 6$

173 Subtract 3:

174 $2x = 3$

175 Divide by 2:

176 $x = \frac{3}{2}$

177 Well explained 👍

178 <h3>Problem 3</h3>

179 Solve: 2x² + 3x - 2 = 0

180 Okay, lets begin

181 $x = \frac {1}{2}, x = -2$

182 <h3>Explanation</h3>

183 First, identify $a = 2, b = 3, c = -2$

184 Using the formula for formula method:

185 $x = {-b \pm \sqrt{b^2-4ac} \over 2a}$

186 $x = \frac{-3 \pm \sqrt{9 + 16}}{4} $

187 $x = \frac{-3 \pm \sqrt{25}}{4} $

188 $x = \frac{-3 \pm 5}{4} $

189 $x = \frac{2}{4} = \frac{1}{2} \quad \text{or} \quad x = \frac{-8}{4} = -2 $

190 Well explained 👍

191 <h3>Problem 4</h3>

192 Solve: √(x + 5) = 6

193 Okay, lets begin

194 $x = 31$

195 <h3>Explanation</h3>

196 We first square both sides:

197 $\left(\sqrt{x + 5}\,\right)^2 = 6^2 $

198 $x + 5 = 36$

199 Now, subtract 5 to isolate x:

200 $x = 31$

201 Well explained 👍

202 <h3>Problem 5</h3>

203 Solve: 2(x - 4) = 10

204 Okay, lets begin

205 $x = 9$

206 <h3>Explanation</h3>

207 Expanding the brackets:

208 $2x - 8 = 10$

209 Add 8 on both sides: $2x = 18$

210 Divide by 2 to isolate x:

211 $\frac {2x}{2} = \frac {18}{2}$

212 $x = 9$

213 Well explained 👍

214 <h2>FAQs on Solving Equations</h2>

215 <h3>1.What do you mean by an equation?</h3>

216 An equation is a mathematical statement that uses the equal sign (=) to demonstrate that two expressions are equal.

217 For instance: x + 5 = 10 x + 5 = 10

218 <h3>2.Why is it necessary to solve equations?</h3>

219 In everyday situations like budgeting, shopping, cooking, traveling, and business computations, solving equations enables us to determine unknown values.

220 <h3>3.What do variables in an equation mean?</h3>

221 Symbols that represent unknown values in an equation are called variables, and they are generally letters like x, y, or z.

222 <h3>4.What does the term “balance” mean in an equation?</h3>

223 It means applying the same operation to both sides of the equation to maintain their equality.

224 <h3>5.How is a simple algebraic equation solved?</h3>

225 By carrying out inverse operations (such as subtraction, addition, multiplication, or division), we can isolate the variable on one side of the equation and solve it.

226 <h3>6.How do I explain to my kid about equations in a simple way?</h3>

227 Use balanced analogy. Tell the that "an equation is like a balance scale. Whatever you do on one side, you must do on the other to keep it balanced."

228 <h3>7.How should I help my child start solving equations?</h3>

229 Start with simple<a>one-step equations</a>. Teach them inverse operations so that they'll not be confused.

230 Addition ↔ subtraction

231 Multiplication ↔ division

232 <h3>8.How can I check if my child’s answer is correct?</h3>

233 Substitute the solution back into the original equation. If the left side of the equation is the same as the right side of the equation, the answer is correct.

234 <h3>9.How can equations be related to real life for kids?</h3>

235 They can be helpful to your kid in many ways, such as:

236 <ol><li>Splitting<a>money</a>or food equally</li>

237 <li>Calculating travel time or speed</li>

238 <li>Adjusting recipes</li>

239 <li>Determining costs, discounts, and profits</li>

240 </ol><h2>Jaskaran Singh Saluja</h2>

241 <h3>About the Author</h3>

242 Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.

243 <h3>Fun Fact</h3>

244 : He loves to play the quiz with kids through algebra to make kids love it.