Evaluate Each Expression

Evaluate each expression is very simple concept in math. When you substitute a particular value for each variable and then act the operations called evaluating expression. The mathematical expression is algebraic, it involves a finite sequence of variables and numbers and then algebraic operations. Evaluation is simplifying the expression.

Algebraic operations are:

Addition

subtraction

multiplication

division

raising to a power

Extracting a root.

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Evaluating algebraic expressions are using PEMDAS method.

Rules for PEMDAS method:

1. First it performs operations inside the parenthesis.

2. Second exponents

3. Third multiplication and division from left to right

4. Finally addition and subtraction from left to right

Evaluate each expression for x=-5 and y=8

2x4-x2+y2

Solution:

Given expression is 2x4-x2+y2

Substitute x and y values in given expression

So,

=2(-5)4-(-5)2+ (8)2

=2(625)-(25) +64

=1250-25+64

=1314-25

=1289

Therefore answer is 1289

Evaluate the expression x3-x2y+9 for x=3, y=9

Solution:

Given expression x3-x2y+9

Plugging the x and y values in given expression

=33-329+9

=27-9*9+9

=27-81+9

=36-81

=-45

Therefore answer is -45

example problems for evaluating the expressions:

Evaluate each expression using PEMDAS method

3 * (9 + 12) - 62 `-:` 2 + 8

Solution:

Remember the PEMDAS rules

First parenthesis: (9+12) = (21)

3*21-62`-:` 2+8

Then exponents: 62 = 6 *6 = 36

3*21-36`-:` 2+8

Then multiplication and division: 3 * 21 = 63 and 36`-:` 2 = 18

63-18+8

Then addition: 63+8=71

71-18

Finally subtract the term

53

Therefore the answer is 53

Evaluate each expression for a = –2, b = 3, c = –4, and d = 4.

bc3 – ad

Solution:

= (3) (–4)3 – (–2) (4)

using PEMDAS exponent rule (-4)3=-4*-4*-4 = -64

= (3) (–64) – (–8)

then multiply the numbers 3*(-64) = -192

= –192 + 8

Add the numbers

= –184

(b + d)2

Solution:

= ((3) + (4))2

Add the inside numbers 3+4=7

= (7)2

using exponent rule 72=7*7=49

= 49

so, the answer is 49

a2b

Solution:

= (–2)2(3)

= (4) (3)

= 12

a – cd

Solution:

= (–2) – (–4)(4)

= (-2)-(-16)

= (-2) +16

= 16-2

= 14

b2 + d2

Solution:

= (3)2 + (4)2

= 9+16

= 25

Evaluate each expression For x = –3.

3x2 – 12x + 4

Solution:

= 3(–2)2 – 12(–2) + 4

= 3(4) + 24 + 4

= 12+24+4

=40

x4 + 3x3 – x2 + 6

Solution:

= (–3)4 + 3(–3)3 – (–3)2 + 6

= 81 + 3(–27) – (9) + 6

= 81 – 81 – 9 + 6

= –3