Please help me
thanks!!


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
[tex]\dfrac{2x + 6}{(x + 2)^2} - \dfrac{2}{(x + 2)}[/tex]
multiply numerator and denominator of the second fraction by (x + 2)
[tex]\dfrac{2x + 6}{(x + 2)^2} - \dfrac{2(x + 2)}{(x + 2)*(x + 2)}[/tex]
Remove the numerator brackets in the right hand fraction. Look out for the minus sign.
[tex]\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}[/tex]
Answer: [tex]\bold{x=\dfrac{21}{4}}[/tex]
Step-by-step explanation:
[tex]\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}}[/tex]
Answer: a = 2
Step-by-step explanation:
[tex].\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}[/tex]