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Two numbers such that their sum is 8 and their difference is also 8?
This question equates to, find x and y such that x + y = 8 x - y = 8 Solving the second equation for x: x = 8 + y Substituting in to the first equation: 8 + y + y = 8 2y = 0 y = 0 Solving for x: x - 0 = 8 x = 8. The two numbers are 8 and 0.
How do you solve x plus y equals 10 and y equals x plus 8?
If x + y = 10 and y = x + 8Substitute y = x + 8 into x + y = 10Which gives; x + x + 8 = 10So; 2x + 8 = 10( - 8 from both sides); 2x=2So: x =1. Substitute back into y = x + 8And therefore y = 1 + 8 ; y=9
-x+ 4 y>-8 x+y8 y?
-x+ 4y>-8, x+y≤3 y≤3-x -x+4(3-x)>-8 -x+12-4x>-8 12-5x>-8 -5x>-20 x>4 y≤3-(4) y≤-1
What is the answer to x- y equals y if x equals 8?
For this problem, write it out... x-y=y... x=2y...8=2y...8/2=y...4=y First, I made both y's on the same side( I added a y to both sides). Then, I replaced the x with an 8. Then, I divide out the 2 from both sides. This leaves me with 4=y So... x=8, y=4, 8-4=4
What is 8 times a number x decreased by y?
It is difficult to tell what expression you are trying to convey. 8 multiplied by x, and then take y away, is 8x-y x take away y, and then 8 multiplied by this number, is 8(x-y) 8 multiplied by x, and divided by y is 8x/y
