In algebraAlgebra is a branch of mathematics that substitutes letters for numbers. Algebra is about finding the unknown or putting real-life variables into equations and then solving them., substitution means replacing letters with numbers. The operation sign is important when carrying out the substitution.
Example
If a = 2, b = -5, c = 10, d = 3
Evaluate:
i. \( \scriptsize a \: + \: b \: + \: c \)
ii. \( \scriptsize c \: \div \: b \)
iii. \( \scriptsize a \: \times\: d \)
iv. \( \frac{d \: \times \: c}{a} \)
v. \( \frac{2b \: \times \: c}{10} \)
Solution
i. \( \scriptsize a \: + \: b \: + \: c \)
From the question the values given are a = 2, b = -5, c = 10
Substituting the values for a, b and c into the equation we get
= \( \scriptsize 2 \: + \: (-5) \: + \: 10 \)
Note:
\( \scriptsize +(-) \; = \; – \)or
\(\scriptsize \; + \; \times \; \; – \; = \; – \)= \( \scriptsize 2 \; – \: 5 \: + \: 10 \)
Using Bodmas rule, Addition comes before subtraction
Add 2 + 10 before subtracting 5
\( \scriptsize 2\: + \: 10 \; – \: 5 \)= \( \scriptsize 12 \; – \: 5 \)
= \( \scriptsize 7\)
ii. \( \scriptsize c \: \div \: b \)
From the question, the values given are c = 10, b = -5
Substituting the values for c and b into the equation we get
\( \scriptsize 10 \: \div \: (-5) \)= \( \frac {10}{-5}\)
= \( \scriptsize \: -2 \)
iii. \( \scriptsize a \: \times\: d \)
From the question, the values given are a = 2, d = 3
Substituting the values for a and d into the equation we get
\( \scriptsize 2 \: \times\: 3 = 6 \)iv. \( \frac{d \: \times \: c}{a} \)
From the question, the values given are a = 2, c = 10, d = 3
Substituting the values for a, c and d into the equation we get
\( \frac{3 \: \times \: 10}{2} \\ \frac{30}{2} \\ = \scriptsize 15\)v. \( \frac{2b \: \times \: c}{10} \)
From the question, the values given are b = -5, c = 10
Substituting the values for b and c into the equation we get
\( \frac{2(-5) \: \times \: 10}{10} \\ = \frac{-10 \: \times \: 10}{10} \\ = \frac{-100}{10} \\ = \scriptsize -10 \)
Responses