Question

2(2x² + xy-z²)-xy. when x=2, y = (-1) and z = (-2)​

Answer 1

`\red{22}`

Explanation:

Given that,

x = 2

y = ( -1 )

z = ( -2 )

To find the value of given expression, you have to replace x, y and z with 2, -1 and -2 respectively.

Let us solve it now.

`2(2x² + xy-z²)-xy \\ 2((2 * 2 ^(2)) + (2 * - 1 ) - ( { - 2}^(2))) - (2 * - 1 )`

You have to solve this equation according to the BODMAS theorem.

That is,

B = brackets

O = of

D = division

M = multiplication

A = addition

S = subtraction

`2((2 * 2 ^(2)) + (2 * - 1 ) - ( { - 2}^(2))) - (2 * - 1 ) \\ 2((2 * 4) - 2 - ( - 4)) - ( - 2) \\ 2(8 - 2 + 4) + 2 \\ 2(6 + 4) + 2 \\ 2 * 10 + 2 \\ 20 + 2 \\ 22`

Answer 2

Answer:
`\Large\boxed{6}`

Explanation:

Given expression

`2(2x^2+xy-z^2)-xy`

Given information

`x=2\\`

`y=-1`

`z=-2`

Substitute values into the given expression

`2[2(2)^2+(2)(-1)-(-2)^2]-(2)(-1)`

Simplify the exponents

`2[2(4)+(2)(-1)-(4)]-(2)(-1)`

Simplify by multiplication

`2[8-2-4]+2`

Simplify values in the parenthesis

`2[6-4]+2`

`2[2]+2`

Simplify by multiplication

`4+2`

`\Large\boxed{6}`

Hope this helps!! :)

Please let me know if you have any questions