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14 answers

1...All the variables power is 1.

2006-12-29 11:34:32 · answer #1 · answered by feanor 7 · 1 0

For challenge a million the first step is to combine like words by including and subtracting from both aspect. 5x -2x = 13 -a million Now upload those 3x = 12 now divide by 3 to get x by itself x = 4 As for challenge 2 commence off a similar way 7x - 2x = 8 + 2 combine the words 5x = 10 Now divide by 5 to get x by itself x = 2

2016-12-01 07:48:44 · answer #2 · answered by ? 4 · 0 0

1- 3x -7y = 2x+8 is linear
it is x-7y -8 =0
2- 2x-8 = y^ 2 is not linear

2006-12-29 12:05:59 · answer #3 · answered by imamulleith 2 · 0 0

A linear equation is one where all terms are to the first power.

If we solve both of these for y, we see that we still have the form x^1; whereas in the second we have x^(1/2).

So the first equation is the linear equation.

2006-12-29 12:07:36 · answer #4 · answered by Anonymous · 0 0

3x-7y=2x+8 is the linear equation

2006-12-29 11:37:05 · answer #5 · answered by Anonymous · 1 0

3x - 7y = 2x + 8
-7y = -x + 8
y = (1/7)x - (8/7) or (x - 8)/7

since this is in the form of y = mx + b, this one is linear.

2x - 8 = y^2
y = sqrt(2x - 8)

This problem isn't linear.

2006-12-29 12:49:39 · answer #6 · answered by Sherman81 6 · 0 0

I assume you mean linear. Equations which result in a graph that is a straight line is called linear. Equations that have exponents equal to 1 and do not have terms like X*Y are linear.
Equation 1 is linear.
Equation 2 is not linear.

2006-12-29 12:08:39 · answer #7 · answered by anonimous 6 · 0 0

3x-7y=2x+8 is linear

2006-12-29 11:35:21 · answer #8 · answered by raj 7 · 1 0

The first one, because all the variables appear
to the first power only. In the second one,
the y is squared and that makes it nonlinear.
By the way, "liner" should be spelled "linear".

2006-12-29 11:44:31 · answer #9 · answered by steiner1745 7 · 1 0

Just the first one, the second one I think would be parabola:

-7y=-x+8
y=1/7x-8/7

2006-12-29 11:45:41 · answer #10 · answered by Anonymous · 0 0

1. 2 is not since y^2 term.

2006-12-29 11:34:30 · answer #11 · answered by I know some math 4 · 1 0

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