Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
Solving a system of equations?
Show step by step how:
Equation 1: x[(1/z) + 1] - y(1/z) - 1/z = y
Equation 2: y[(1/z) + z] - x(1/z) = -x
becomes:
y = [-z + 1]/[z(z^2 + 2z +1)]
Hint: Express equation 2 in terms of x and plug into equation 1.
My attempt at isolating x in equation 2:
y[(1/z) + z] - x(1/z) = -x
y[(1/z) + z] = -x + x(1/z)
y[(1/z) + z] = x[ -1 + (1/z)]
y[(1/z) + z] = x[(1/z) - 1]
y[(1/z) + z]/[(1/z) - 1] = x
I'm going to leave it to Yahoo Answers users to do the rest, it's "just algebra" so you guys can handle it. Thanks.
3 Answers
- ?Lv 71 year ago
Equation 1:
x[(1/z) + 1] - y(1/z) - 1/z = y
Equation 2:
y[(1/z) + z] - x(1/z) = -x
becomes:
y = [-z + 1]/[z(z^2 + 2z +1)]
Hint: Express equation 2 in terms of x and plug into equation 1.
1.
My attempt at isolating x in equation
2:
y[(1/z) + z] - x(1/z) = -x
y[(1/z) + z] = -x + x(1/z)
y[(1/z) + z] = x[ -1 + (1/z)]
y[(1/z) + z] = x[(1/z) - 1]
y[(1/z) + z]/[(1/z) - 1] = x
- davidLv 71 year ago
Equation 2: y[(1/z) + z] - x(1/z) = -x <<< use the 'hint' ... solve fr x
y[(1/z) + z] = -x + x(1/z)
x[(1/z) - 1] = y[(1/z) + z]
x = y[(1/z) + z] / [(1/z) - 1]
===== now plug into eq 1
Equation 1: x[(1/z) + 1] - y(1/z) - 1/z = y
y[(1/z) + z] / [(1/z) - 1][(1/z) + 1] - y(1/z) - 1/z = y
... from here do a lot of algebra steps t solve for y
It is just algebra .. I leave the rest for you
- ?Lv 71 year ago
So, start by applying their hint. Isolating x in the 2nd eqs gives x = ?
Answer that & we will proceed.
OK. Now just plug it into the 1st eqs to get
y[(1/z) + z] / [(1/z) - 1][(1/z) + 1] - y(1/z) - 1/z = y
Now, just work at it to get it to the desired form.
Start by isolating y. A little more simplification leads to the desired form.