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drek231 [11]
3 years ago
14

-6x - 2x = 40. Calculate and show work

Mathematics
2 answers:
Roman55 [17]3 years ago
8 0
Solve any algebraic expression by getting the x on one side (usually the left-hand side) and everything else and the opposite side (usually the right-hand side).

We do this by using the inverse operation of what is being done to the variable. If something is adding to the variable, say like an expression like x + 10, then we use subtraction to get rid of the addition and vice versa.

Similar to multiplication, we use division to cancel out the multiplication and vice versa.

If there are any like terms, we combine them. I will assume that you know how to combine like terms, but in short, to combine like terms, add the coefficients together and use the common variables and power. If there is no coefficient, then a 1 is implied. Similarly, if no explicit power is shown, then a 1 is implied.

-6x - 2x = 40
-8x        = 40

-6x and -2x are like terms because they have the same variable and are to the same power.

-6x + (-2)x
(-6 - 2)x
-8x

Now we have the expression -8x = 40. Now we can use division to get rid of the -8 juxtaposed to the variable x.

-8x = 40
-8x / 8 = 40 / -8
x = -5

xxMikexx [17]3 years ago
3 0
-6x-2x= 40
-8x=40
X= 40/-8
X= -5
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Use Gauss-Jordan elimination to solve the following linear system.
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Just put the coefients in to a matrix

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\left[\begin{array}{ccc}1&-6&-3|4\\-2&0&-3|-8\\-2&2&-3|-14\end{array}\right]
mulstiply 2nd row by -1 and add to 3rd
\left[\begin{array}{ccc}1&-6&-3|4\\-2&0&-3|-8\\0&2&0|-6\end{array}\right]
divde last row by 2
\left[\begin{array}{ccc}1&-6&-3|4\\-2&0&-3|-8\\0&1&0|-3\end{array}\right]
multiply 2rd row by 6 and add to top one
\left[\begin{array}{ccc}1&0&-3|-14\\-2&0&-3|-8\\0&1&0|-3\end{array}\right]
multiply 1st row by -1 and add to 2nd
\left[\begin{array}{ccc}1&0&-3|-14\\-3&0&0|6\\0&1&0|-3\end{array}\right]
divide 2nd row by -3
\left[\begin{array}{ccc}1&0&-3|-14\\1&0&0|-2\\0&1&0|-3\end{array}\right]
mulstiply 2nd row by -1 and add to 1st row
\left[\begin{array}{ccc}0&0&-3|-12\\1&0&0|-2\\0&1&0|-3\end{array}\right]
divide 1st row by -3
\left[\begin{array}{ccc}0&0&1|4\\1&0&0|-2\\0&1&0|-3\end{array}\right]

rerange
\left[\begin{array}{ccc}1&0&0|-2\\0&1&0|-3\\0&0&1| 4\end{array}\right]

x=-2
y=-3
z=4
(x,y,z)
(-2,-3,4)

B is answer
7 0
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