Question

Please help me thanks!!

Answer 1

Answer:

Step-by-step explanation:

Question One

Multiply through by 2

2*1/2 * (2x + y ) = 21/2 * 2

Combine

2x + y = 21

Subtract 2x from both sides

y = 21 - 2x

Now equate the two given equations

y = 21 - 2x

y = 2x

Add 2x to both sides

2x = 21 - 2x

2x + 2x = 21

4x = 21

x = 21/4

x = 5 1/4

or

x = 5.25

Question 2

$\dfrac{2x + 6}{(x + 2)^2} - \dfrac{2}{(x + 2)}$

multiply numerator and denominator of the second fraction by (x + 2)

$\dfrac{2x + 6}{(x + 2)^2} - \dfrac{2(x + 2)}{(x + 2)*(x + 2)}$

Remove the numerator brackets in the right hand fraction. Look out for the minus sign.

$\dfrac{2x + 6}{(x + 2)^2} - \dfrac{2(x + 2)}{(x + 2)^2}\\\\\dfrac{2x + 6- 2x - 4}{(x + 2)^2}}\\\\\dfrac{2}{(x + 2)^2}$

Answer 2

Answer:  $\bold{x=\dfrac{21}{4}}$

Step-by-step explanation:

$\dfrac{1}{2}(2x+y)=\dfrac{21}{2}\\\\\text{Multiply both sides by 2 to clear the denominator:}\\2x + y = 21\\\\\text{Now, the system is:}\bigg\{{2x+y=21\atop{y=2x}}\\\\\text{Substitute y in the first equation with 2x to solve for x:}\\2x + y = 21\\2x + 2x = 21\\.\qquad 4x=21\\\\.\qquad \large\boxed{x=\dfrac{21}{4}}$

Answer:  a = 2

Step-by-step explanation:

$.\quad \dfrac{2x+6}{(x+2)^2}-\dfrac{2}{x+2}\bigg(\dfrac{x+2}{x+2}\bigg)\\\\\\=\dfrac{2x+6}{(x+2)^2}+\dfrac{-2(x+2)}{(x+2)^2}\bigg\\\\\\\\=\dfrac{2x+6-2x-4}{(x+2)^2}\\\\\\=\dfrac{2}{(x+2)^2}\implies \large\boxed{a=2}$