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rusak2 [61]
3 years ago
6

Solve the system by substitution. y=2x y=10x-16

Mathematics
2 answers:
anygoal [31]3 years ago
8 0

Answer:

y=4  x=2

Step-by-step explanation:

y=2x ⇒ I

y=10x-16 ⇒ II

y/2=2x/2

x=y/2 OR 1/2(y)

substitute I in II

y=10(y/2)-16

y=5y-16

SWAP -16 AND y

5y-y=16

4y/4=16/4

y=4

y=2x

(4)/2=2x/2

x=2

Debora [2.8K]3 years ago
4 0
You would first plug in y to the other equation. It doesn’t matter which one you do. So it would look like...
2x=10x-16
Then you would move the like terms to one side
2x-10x=-16
-8x = 16
Then make the x alone by dividing -8 on both sides to solve for x.
X= 16/-8
x = -2
Hope this helped!


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Step-by-step explanation:

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Which matrix represents the system of equations shown below? 3x-5y=12 4x-2y=15
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Answer:

\left[\begin{array}{ccc}3&-5  &|12\\4&-2  &|15\\\end{array}\right]

Step-by-step explanation:

When making a matrix of two equations with the variables x and y, the result will be a matrix with three columns:

  • a column for the values of x in each equation
  • a column for the values of y in each equation
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since our system of equations is:

3x-5y=12\\ 4x-2y=15

we can see that the value for x in the first equation is 3 and in the second equation is 4, thus the first column will have the numbers 3 and 4:

\left[\begin{array}{ccc}3&&\\4&&\\\end{array}\right]

Now for the values of y we hvae -5 in the first equation and -2 in the second equation, we update the matrix with another column with the values of -5 and -2:

\left[\begin{array}{ccc}3&-5&\\4&-2&\\\end{array}\right]

Finally, the last column is the independent values of each equation (or the results) in the first equation that number is 12 and in the second equation is 15, thus the matrix is:

\left[\begin{array}{ccc}3&-5&12\\4&-2&15\\\end{array}\right]

usually there is a line separating the columns for the values of x and y, and the independent values:

\left[\begin{array}{ccc}3&-5  &|12\\4&-2  &|15\\\end{array}\right]

this is the matrix of the system of equations

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