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Algebra - Substitution into Formulas

Grade 7IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Substitution is the algebraic process of replacing a variable (a letter) with a specific numerical value. Imagine the variable as an empty placeholder box that you fill with a number to calculate a result.

Always use parentheses when substituting values, especially negative numbers. For example, if x=2x = -2, then x2x^2 should be written as (2)2=4(-2)^2 = 4 rather than 22=4-2^2 = -4. This prevents common sign errors in calculations.

The Order of Operations (BIDMAS/BODMAS) must be followed after substitution. Think of this as a sequence: first solve Brackets, then Indices (powers/roots), then Division and Multiplication from left to right, and finally Addition and Subtraction.

When a variable is written directly next to a number or another variable, such as 3x3x or abab, it implies multiplication (3×x3 \times x or a×ba \times b). Visualize these terms as being 'glued' together until they are evaluated.

Fractional formulas require you to treat the numerator and denominator as separate groups. In a formula like a+bc\frac{a + b}{c}, visualize a horizontal bar acting as a divider; you must calculate the entire top value (a+ba + b) before dividing by cc.

Real-world formulas describe physical relationships. For example, in the perimeter formula P=2l+2wP = 2l + 2w, you can visualize a rectangle where you add two lengths (ll) and two widths (ww) to find the total boundary length.

Powers and Roots: If a variable is raised to a power, such as y3y^3, substitute the number first and then multiply it by itself the required number of times. If y=2y = 2, then y3y^3 becomes 2×2×2=82 \times 2 \times 2 = 8.

📐Formulae

P=2(l+w)P = 2(l + w)

A=12bhA = \frac{1}{2}bh

C=59(F32)C = \frac{5}{9}(F - 32)

v=u+atv = u + at

d=rtd = rt

y=mx+cy = mx + c

💡Examples

Problem 1:

Given the formula v2=u2+2asv^2 = u^2 + 2as, find the value of v2v^2 when u=5u = 5, a=3a = 3, and s=4s = 4.

Solution:

Step 1: Substitute the given values into the formula: v2=(5)2+2(3)(4)v^2 = (5)^2 + 2(3)(4). \ Step 2: Calculate the power: 52=255^2 = 25. \ Step 3: Calculate the product: 2×3×4=242 \times 3 \times 4 = 24. \ Step 4: Add the results: 25+24=4925 + 24 = 49.

Explanation:

First, replace each letter with its value using parentheses. Following the order of operations, evaluate the square of uu and the multiplication of 22, aa, and ss separately before adding them together.

Problem 2:

Evaluate the expression z=3xywz = \frac{3x - y}{w} when x=4x = 4, y=2y = -2, and w=7w = 7.

Solution:

Step 1: Substitute the values: z=3(4)(2)7z = \frac{3(4) - (-2)}{7}. \ Step 2: Multiply in the numerator: 3×4=123 \times 4 = 12. \ Step 3: Simplify the double negative in the numerator: 12(2)=12+2=1412 - (-2) = 12 + 2 = 14. \ Step 4: Divide by the denominator: z=147=2z = \frac{14}{7} = 2.

Explanation:

Substitute the values carefully, noting that subtracting a negative number becomes addition. Calculate the entire numerator first to get a single number before performing the final division by ww.