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So, first we multiply the fraction by using the formula a/b times c/d= a times c/b times d
=(y^2-16) times 5y/2y(y-4)
Now, we cancel the common factor y
=(y^2-16) times 5/2(y-4)
Now, we factor 5(y^2-16)
We factor (y^2-16) first
y^2-16
Rewrite 16 as 4^2
y^2-4^2
Now, apply the formula x^2-y^2=(x+y)(x-y)
=y^2-4^2=(y+4)(y-4)
=5(y+4)(y-4)
=5(y+4)(y-4)/2(y-4)
Cancel the common factor y-4
=5(y+4)/2
Answer: 5(y+4)/2
Answer:
(1,2) is a solution to both equations
Step-by-step explanation:
To determine if (1,2) is a solution to both equations, substitute into the equation and see if it is true
2x+y =4
2(1) + 2 =4
2+2 =4
4=4
true
y =3x-1
2 = 3(1) -1
2 =3-1
2=2
true
Since both statements are true
(1,2) is a solution to both equations
Answer:
The answer is 4
Step-by-step explanation:
Given,

Now, the expression
∴
hope you have understood this...
pls mark my answer as the brainliest
H= 2a/a+b woolud be the answer