Substitution into Algebraic Expressions

To substitute means to replace letters in algebraic expressions with values. 

Example 2.3.1:

Find the value of

(a) \( \scriptsize 4x \)  
(b) \( \scriptsize xy \: – \: 5y \)  
(c) \( \scriptsize 5x^2 \: – \: 2y^3 \)  

When x = 2, and y = 3 

Solution 

a. \( \scriptsize 4x \\ = \scriptsize 4 \: \times \: x \\ \scriptsize (x = 2) \\ = \scriptsize 4 \: \times \: 2 = \scriptsize 8\)  

Note: We replaced/substituted x with 2

b. \( \scriptsize xy \: – \: 5y \)  

= \( \scriptsize (x \: \times \: y) \: – \: (5 \: \times \: y) \)  

When x = 2, and y = 3 

Substitute the values into the equation

= \( \scriptsize (2 \: \times \: 3) \: – \: (5 \: \times \: 3) \)  

= \( \scriptsize 6 \: – \: 15 \)  

= \( \scriptsize -9 \)  

c. \( \scriptsize 5x^2 \: – \: 2y^3 \)

= \( \scriptsize (5 \: \times \: x \: \times \: x ) \: – \: (2 \: \times \: y \: \times \: y \: \times \: y)\)

When x = 2, and y = 3 

Substitute the values into the equation

= \( \scriptsize (5 \: \times \: 2 \: \times \: 2 ) \: – \: (2 \: \times \: 3 \: \times \: 3 \: \times \: 3)\)

= \( \scriptsize 20 \: – \: 54 = \; -34\)

Example 2.3.2: 

If x = -4, y = 5, z = \( \frac{1}{2}\), m = 10 and n = 0

Evaluate the following; 

a. \( \scriptsize zn \: + \: xyz \)

b. \( \frac {(-3x \: – \: ym)} {2m \: – \: y^2 }\)

Solution 

a. \( \scriptsize zn \: + \: xyz \)

= \( \scriptsize ( z \: \times \: n) \: + \: (x\: \times \:y\: \times \:z) \)

If x = -4, y = 5, z = \( \frac{1}{2}\), m = 10 and n = 0

Substitute the values into the equation

= \(\normalsize \frac{1}{2} \scriptsize \: \times \: 0 \: + \: \left (-4 \: \times \: 5\: \times \: \normalsize \: – \frac{1}{2} \right)\)

= \(\scriptsize 0 \: + \: \left ({\color{Red} -}4 \: \times \: 5\: \times \: {\color{Red} -} \normalsize \frac{1}{2} \right)\)

\( \scriptsize {\color{Red} -} \: \times \: {\color{Red} -} = {\color{Blue} +} \)

= \(\scriptsize 0 \: + \: \left (-4 \: \times \: – \normalsize \frac{1}{2} \scriptsize \: \times \: 5 \right)\)

= \(\scriptsize 0 \: + \: \left (+2 \: \times \: 5 \right)\)

= \(\scriptsize 0 \: + \: \left (\scriptsize + 10 \right)\)

= \(\scriptsize 0 \: + \: 10 \)

= \(\scriptsize 10 \)

b. \( \frac {(-3x \: – \: ym)} {2m \: – \: y^2 }\)

= \( \frac {(-3\: \times \: x) \: – \: (y\: \times \:m)} {(2\: \times \:m) \: – \: (y \: \times \: y) }\)

If x = -4, y = 5, z = \( \frac{1}{2}\), m = 10 and n = 0

Substitute the values into the equation

= \( \frac {\left [(-3\: \times \: -4) \: – \: (5\: \times \:10) \right]} {(2\: \times \:10) \: – \: (5 \: \times \: 5) }\)

= \( \frac{ -(-12 \: – \: 50)}{20 \: – \: 25} \)

= \( \frac{ {\color{Red} -}({\color{Red} -}62)}{-5} \)

\( \scriptsize {\color{Red} -} \: \times \: {\color{Red} -} = {\color{Blue} +} \)

= \( \frac{{\color{Blue} +}62}{{\color{Red} -}5} \)

\( \scriptsize {\color{Blue} +} \: \div \: {\color{Red} -}\: = \: {\color{Red} -} \)

= \({\color{Red} -} \frac{62}{5} \)

= \( \scriptsize -12 \frac{2}{5} \)

Evaluation Questions: