Question

A. 13а - 5 +b, при а = 2, b = 11 b. 21-x+2y, прих = 13, y= 3101x + 101у, при х = 2, y=-2d. 191 - (3x+3y), при х = 2, y= 10е. 191 — 3(x+y), прих = 2, y = 10f. 53 + 4а — 4b, при а = 2, b = 1g. 53 + 4(a-b), при а=2, b=1​

Answer 1

a. 13а - 5 +b, при а = 2, b = 11
13•2 - 5 + 11 =
= 26 - 5 + 11 =
= 37 - 5 = 32

b. 21-x+2y, прих = 13, y= 3
21 - 13 + 2•3 =
= 21 - 13 + 6 =
= 27 - 13 = 14

c. 101x + 101у, при х = 2, y=-2
101•2 + 101•(-2) =
= 202 - 202 = 0

d. 191 - (3x+3y), при х = 2, y= 10
191 - (3•2 + 3•10) =
= 191 - (6 + 30) =
= 191 - 36 = 155

е. 191 — 3(x+y), прих = 2, y = 10
191 - 3•(2 + 10) =
= 191 - 3•12 =
= 191 - 36 = 155

f. 53 + 4а — 4b, при а = 2, b = 1
53 + 4•2 - 4 • 1 =
= 53 + 8 - 4 =
= 61 - 4 = 57

g. 53 + 4(a-b), при а=2, b=1
53 + 4•(2 - 1) =
= 53 + 4•1 =
= 53 + 4 = 57

Answer 2

Let's substitute the given values of $a$ and $b$ into the given expressions to calculate the results:

a. $13a - 5 + b$ when $a = 2$ and $b = 11$ $= 13(2) - 5 + 11$ $= 26 - 5 + 11$ $= 32$

b. $21 - x + 2y$ when $x = 13$ and $y = 3101(2) + 101(1)$ $= 21 - 13 + 2(6202 + 101)$ $= 21 - 13 + 12405 + 202$ $= 12515$

c. $191 - (3x + 3y)$ when $x = 2$ and $y = 10$ $= 191 - [3(2) + 3(10)]$ $= 191 - [6 + 30]$ $= 191 - 36$ $= 155$

d. $191 - 3(x + y)$ when $x = 2$ and $y = 10$ $= 191 - 3(2 + 10)$ $= 191 - 36$ $= 155$

e. $53 + 4a - 4b$ when $a = 2$ and $b = 1$ $= 53 + 4(2) - 4(1)$ $= 53 + 8 - 4$ $= 57$

f. $53 + 4(a - b)$ when $a = 2$ and $b = 1$ $= 53 + 4(2 - 1)$ $= 53 + 4(1)$ $= 57$

So, the results for the given expressions with the provided values of $a$ and $b$ are:

a. $32$ b. $12515$ c. $155$ d. $155$ e. $57$ f. $57$