Solving Linear Equations
[latex]2x + 3 = 7[/latex]
Learning level: Grade school
Distinguish between an algebraic expression and an algebraic equation
what you need to know
An algebraic expression is a combination of variables, constants, and operations, without an equal sign, representing a value. An algebraic equation, on the other hand, includes an equal sign and is used to find the value of the variable(s) that satisfy the equation.
Algebraic Expression:
\[3x^2 + 2y - 5\]
This is an algebraic expression because it consists of terms involving variables \(x\) and \(y\), along with constants and operations. It is not stating that the expression is equal to anything; it is just a representation of a mathematical expression.
Algebraic Equation
\[4x + 7 = 15\]
This is an algebraic equation since it has an equal sign and the aim is to find the value of the variable \(x\) that satisfies the equation.
Worked Example
Given the equation:
\[2x + 5 = 11\]
Let's solve this equation to find the value of \(x\).
Subtract 5 from both sides:
\[2x + 5 - 5 = 11 - 5\]
\[2x = 6\]
Divide by 2:
\[\frac{2x}{2} = \frac{6}{2}\]
\[x = 3\]
Therefore, the solution to the equation \(2x + 5 = 11\) is \(x = 3\).
Solving linear equations with one unknown, including those involving brackets and fractions
what you need to know
- Basic Linear Equations: These are straightforward equations like \(2x + 3 = 7\), where you isolate \(x\) by performing inverse operations (subtracting, dividing, etc.) on both sides of the equation until \(x\) is alone on one side.
- Equations with Brackets: Equations such as \(3(x + 2) - 2 = 10\) involve distributing or expanding the terms within the brackets first, before simplifying the equation further.
- Equations with Fractions: Equations like \(\frac{2x}{3} + \frac{x - 1}{2} = 4\) require clearing fractions by multiplying through by the least common denominator of all fractions involved.
- Combination of Brackets and Fractions: Equations such as \(2(3x - \frac{1}{2}) = 5\) require both distribution and handling of fractions in the process of solving for \(x\).
Note: Solving linear equations with one unknown, whether they involve brackets, fractions, or both, requires systematic application of algebraic rules and operations to isolate the variable and determine its value.
Practical problems involving simple linear equations
Practical problems involving simple linear equations often require translating a real-world scenario into an algebraic equation that can be solved to find the unknown value. Here’s how we approach and differentiate these problems:
what you need to know
Basic Linear Equation Problems:
- Example: If you buy 3 notebooks for $15, how much does one notebook cost?
- Translation to Equation: \(3x = 15\)
- Solution: Divide both sides by 3 to find \(x = 5\). So, each notebook costs $5.
- Example: If you buy 3 notebooks for $15, how much does one notebook cost?
Problems Involving Brackets:
- Example: A rectangle has a length that is 3 units more than twice its width. If the perimeter is 24 units, what are the dimensions?
- Translation to Equation: Let ww be the width. The length is 2w+32w + 3. The perimeter equation is 2(w+2w+3)=242(w + 2w + 3) = 24.
- Solution: Simplify and solve 2(3w+3)=242(3w + 3) = 24, leading to 3w+3=123w + 3 = 12, then 3w=93w = 9, and finally w=3w = 3. The width is 3 units, and the length is 2(3)+3=92(3) + 3 = 9 units.
- Example: A rectangle has a length that is 3 units more than twice its width. If the perimeter is 24 units, what are the dimensions?
Problems Involving Fractions:
- Example: If you save 13\frac{1}{3} of your allowance every week and you saved $50 in 5 weeks, what is your weekly allowance?
- Translation to Equation: Let aa be the weekly allowance. Each week, you save 13a\frac{1}{3}a. Over 5 weeks, 5⋅13a=505 \cdot \frac{1}{3}a = 50.
- Solution: Simplify and solve 53a=50\frac{5}{3}a = 50, leading to 5a=1505a = 150, then a=30a = 30. The weekly allowance is $30.
- Example: If you save 13\frac{1}{3} of your allowance every week and you saved $50 in 5 weeks, what is your weekly allowance?
Combination of Brackets and Fractions:
- Example: You have $100 to spend on t-shirts. Each t-shirt costs $x and you also have to pay a fixed shipping fee of 14\frac{1}{4} of the total cost. How many t-shirts can you buy if you spend all your money?
- Translation to Equation: Let nn be the number of t-shirts. The total cost is nxnx, and the shipping fee is 14nx\frac{1}{4}nx. The equation is nx+14nx=100nx + \frac{1}{4}nx = 100.
- Solution: Combine terms to get 54nx=100\frac{5}{4}nx = 100, then 5nx=4005nx = 400, leading to nx=80nx = 80. Solve for nn to find the number of t-shirts you can buy given the price xx.
- Example: You have $100 to spend on t-shirts. Each t-shirt costs $x and you also have to pay a fixed shipping fee of 14\frac{1}{4} of the total cost. How many t-shirts can you buy if you spend all your money?
Translating verbal statements into simple linear equations
Translating verbal statements into simple linear equations involves converting a written description of a problem into an algebraic equation that can be solved. Here are the steps and examples to help understand the process
