Do you mean y=x^2.5? I you had y=13^2.77, it's easier to use log.
log y=2.77*log13 ~ 3.0856. 1217.9 is the antilog and answer.x=1217.9
But math can be more complicated. How about y^2.5=x^1.8. Logs really shine here. Take log of both sides. 2.5*log y = 1.8 log x. Say x=100 and 1.8 log 100 = 1.8*2=3.6. We have 2.5 log y = 3.6 and log y = 3.6/2.5 = 1.44. Now y = antilog 1.44=27.54229. So does 27.54229^2.5 = 100^1.8 ? Yes it does.
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You multiply one or both equations by some constant (especially chosen for the next step), and add the two resulting equations together. Here is an example: (1) 5x + 2y = 7 (2) 2x + y = 3 Multiply equation (2) by -2; this factor was chosen to eliminate "y" from the resulting equations: (1) 5x + 2y = 7 (2) -2x -2y = -6 Add the two equations together: 3x = 1 Solve this for "x", then replace the result in any of the two original equations to solve for "y".
Algebraic equations with two variables will need two equations to be able to solve it. Then, you can solve it with either substitution, adding/subtracting them together, or graphing! Those are the basic steps... For example: An instance of substitution: 2x + 1 = y + 2 x + y = 3 You could isolate y in the second equation to equal y = 3-x. Then in the first equation, substitute y with what it equals to 2x + 1 = 3-x+2 Then you can solve for x!
just use the PEMDAS system. p-parenthesis e-exponents m-multiplication d-division a-adding s-subtracting
You solve one of the equation for one of the variables. For example, if the variables involved are "x" and "y", you might solve for "y". It doesn't really matter what variable you solve for first, so you can solve for whatever variable is easiest to solve. Then - assuming you got, for example, "y = 3x -1", in this example you would replace every "y" by "3x - 1" in the other equation or equations.
You can use a graph to solve systems of equations by plotting the two equations to see where they intersect