Answer:
using identity (a+b)(a-b) = a^2 -b^2
You need to solve this "system of linear equations." In other words, find a point (x,y) that satisfies both 4x-3y=17 and 2x-5y=-11.
Try solution by elimination. Multiply the 2nd equation by -2 to obtain -4x+5y=22. Add this result to the 1st equation. I'd suggest you write this out to see what is happening.
4x-3y=17
-4x+10y=22
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7y=39. Solving for y, we get y=39/7 (a rather awkward fraction).
Now find x. To do this, substitute 39/7 for y in either of the given equations. Solve the resulting equation for x.
Write your solution in the form (x, y): ( ? , 39/7).
Answer:
x^0 y^-3 / x^2 y^-1
= 1 / x^2 y^-1 (y^3) ...because x^0 = 1 and [(y^-1) (y^3)] = y^2
= 1/(x^2 y^2)
Answer:
802
Step-by-step explanation:
Answer:
180
Step-by-step explanation:
