Can someone please help me with these 2 math questions

1. if x² + y² = 14 and xy = 5, find the value of (x+y)²

2. Given that a + b = 10 and a² - b² = 40, find the value of a - b

1.(x+y)^2 = x^2+xy+xy+y^2

= x^2+2xy+y^2
=14 + 2(5)
= 24

2. this is pretty simple so you can trial and error..like 8+2=10 does 8^2-2^2=40?
no it doesn't so you can try again but doesnt take long cos they gave you simple numbers
a=7
b=3

a+b=10 7+3=10
a^2-b^2= 40 7^2-3^2= 40
they are the same so a-b would = 7-3 = 4

1. Using the equation for the expansion of a perfect square

(x+y)^2 = x^2 + 2xy + y^2
(x+y)^2 = (x^2 + y^2) + 2(xy)

By the information given, you can then say

(x+y)^2 = 14 + 2*5
(x+y)^2 = 24

2. Using the difference of perfect squares law

(a+b)(a-b)=a^2-b^2

By the information given

10(a-b)=40
a-b=40/10
a-b=4

1. solve the second given equation for y to find y=5/x and plug that into the first one to find
x^2+(5/x)^2=14
x^2+25/x^2=14
x^4+25=14x^2
x^4-14x^2+25=0 replace temporarily x^2 with u:
u^2-14u+25=0
with a calculator u then equals 11.89897948, 2.101020514, so since x is ±sqrt(u), x=
±3.44948974, ±1.44948974

y then, being 5/x, must equal ±1.44948974, ±3.44948975 (yes, the same as x)

since squaring x+y will render its sign irrelevant, you can narrow down the options for x+y to (rounded) 2.8, 0, 4.8, and 6.8, and thus use a calculator to find (x+y)^2=8.40408203,0,24,47.5959181

2. solve the first equation to get b=10-a and plug it in to the second to get a^2-(10-a)^2=40
a^2-100+20a-a^2=40
20a-100=40
20a=140
a=7

if a=7 then 7+b=10 and b=3, and a-b is thus 4

(x+y)^2 = x^2 + 2xy + y^2
therefore (x+y)^2 = 14 (x^2 + y^2) + 2*5 (2xy)
= 24

if a+b=10
then a=10-b
put into other equation
(10-b)^2 - b^2 = 40
(100 - 20b +b^2) - b^2 = 40
therefore 100 - 20b = 40
-20b = -60
therefore b=3

put this into a+b=10
a+3=10 therefore a=7
therefore a-b= 7-3
=4

1. if x² + y² = 14 and xy = 5, find the value of (x+y)²
=> x² + y² + 2xy = 24 => the value of (x+y)² = 24.

a=7 and b=4 so 7 take 4=3

sorry cant work 1st one out