i have an algebra question?

4x-y=3 (i)
9x-2y=10 (ii)
now from (i) get the value of either x or y
i take the value of x here
4x=3 + y
=>x=(3 + y)/4
now,put the value of x in (ii)
9(3 + y)/4 - 2y=10
=>(27 + 9y)/4 - 2y=10
=>(27 + 9y - 8y)/4=10
=>27 + y=10(4)
=>27 + y=40
=>y=40 - 27
=>y=13
now put the value of y in any equation
i put it in (i)
4x - (13)=3
=>4x - 13=3
=>4x=3 + 13
=>4x =16
=>x=16/4
=>x=4
therefore
x=4
y=13

For increasing cubic binomials the final formulation is as follows: (a + b) ^ 3 = a^3 + 3*a^2*b^a million + 3*a^a million*b^2 + b^3 on your case, a is x and b is -y^5 So (x - y^5)^3 = x^3 + 3*x^2*(-y^5)^a million + 3*x^a million*(-y^5)^2 + (-y^5)^3 Simplified: =x^3 - 3x^2*y^5 + 3x*y^10 - y^15 :D

Let 4x-y=3 be equation number 1& 9x-2y=10 be equation number 2
from 1 we get 4x=3+y
=>y=4x-3 ................(eq^n3)
substituting this value in 2 we get
9x-2(4x-3)=10
=>9x-8x+6=10
=>x=10-6=4
=>x=4
substituting the value of x to get y in3
y=4(4)-3=16-3=13

4x - y = 3 equation 1
9x - 2y = 10 equation 2

Solve for y in equation 1:

4x - 3 = y

Plug this into equation 2:

9x - 2(4x - 3) = 10

Solve for x:

9x - 8x + 6 = 10
x + 6 = 10
x = 4

Plug this into equation 1 to solve for y:

4(4) - y = 3
16 - y = 3
16 - 3 = y
13 = y

These questions are so fun and easy!
Let's see if i can remember from last year.
y= 4x-3

9x-2(4x-3)=10
9x-8x+6=10
x+6=10
x=4

y=4(4)-3
y=16-3
y=13

y=13
x=4
(4,13)

It's not really that hard! Also, you should do them by yourself

Edit: The guy above me did it wrong he forgot to change to - to a + when multiplying a - and a -.
So it screwed the whole problem.

Let y=4x-3
Substitute y=4x-3 into 9x-2y=10

9x-2(4x-3)=10
solved x=-4

Substitute x=-4 to y=4x-3
solved y=-19

So x=-4 and y=-19
Hope it helps

Sorry, my bad, answer should be x=4, y=13.