Question

PLEASE HELP Match the pairs of equivalent expressions

Answer 1

Answer:

$(4t-\dfrac{8}{5})-(3-\dfrac{4}{3}t)---------------\dfrac{16t}{3}-\dfrac{23}{5}$

$5(2t+1)+(-7t+28)--------------3t+33$

$(\dfrac{-9}{2}t+3)+(\dfrac{7}{4}t+33)-----------------\dfrac{-11t}{4}+36$

$3(3t-4)-(2t+10)-------------7t-22$

Step-by-step explanation:

a)

$(4t-\dfrac{8}{5})-(3-\dfrac{4}{3}t)$

Firstly we will open the parentheses term and then on combining the like terms and then solving them.

$=4t-\dfrac{8}{5}-3+\dfrac{4}{3}t\\\\i.e.\\\\=4t+\dfrac{4}{3}t-\dfrac{8}{5}-3\\\\=\dfrac{4t\times 3+4t}{3}+\dfrac{-8-3\times 5}{5}\\\\=\dfrac{16t}{3}-\dfrac{23}{5}$

b)

$5(2t+1)+(-7t+28)\\\\=5\times 2t+5\times 1-7t+28\\\\=10t+5-7t+28\\\\=10t-7t+5+28\\\\=3t+33$

( Here we used distributive property for first bracket and open the parentheses term and them we combined the like terms and finally simplified our terms )

c)

$(\dfrac{-9}{2}t+3)+(\dfrac{7}{4}t+33)$

We firstly open our parentheses term but there will be no change in the terms since the sign before parentheses is positive.

and then we will combine the like terms and simplify them.

$=\dfrac{-9}{2}t+3+\dfrac{7}{4}t+33\\\\\\=\dfrac{-9}{2}t+\dfrac{7}{4}t+3+33\\\\\\=\dfrac{-9t\times 2+7t}{4}+36\\\\\\=\dfrac{-11t}{4}+36$

d)

$3(3t-4)-(2t+10)$

i.e. we use the distributive property for first bracket and open second parentheses term. since the sign before the second parentheses is negative hence the sign of the terms will get interchanged.

i.e.

$=3\times 3t+3\times (-4)-2t-10\\\\i.e.\\\\=9t-12-2t-10\\\\=9t-2t-12-10\\\\=7t-22$

Answer 2

5 (2t + 1) + (-7t + 28) = 3t + 33

(4t - 8/5) - (3 - 4/3t) = 16/3t - 23/5

3 (3t - 4) - (2t + 10) = 7t - 22

(-9/2x + 3) + (7/4x + 33) = -11/4x + 36