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3 years ago

Solving systems by substitutions

1 answer:

8 0

Let's take two different equations.

4x+2y=8
7x+3y=-10
(I just came up with these on the spot, so the answers might be strange but i should be able to explain nevertheless)
First, let's solve the first equation for "y".
4x+2y=8
-4x -4x
2y=8-4x
/2
y=4-2x
Now that we have "y", let's plug it into the second equation.
7x+3(4-2x)=-10
7x+12-6x=-10
x+12=-10
-12 -12
x=-22
Alright, now we have "x" definitively, let's plug it in to find "y".
4(-22)+2y=8
-88+2y=8
+88 +88
2y=96
/2
y=48
So from this example you can see. You solve one variable in terms of another (solve y as being 4-2x), plug it into the second equation to find one variable's value, then plug that variable's value back in again later to get the first variable's value.

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The augmented matrix of a system of equations has been transformed to an equivalent matrix inâ€‹ row-echelon form. Usingâ€‹ x, y

jenyasd209 [6]

Answer: The system of equations is:

x + 2y + 2 = 4

y - 3z = 9

z = - 2

The solution is: x = -22; y = 15; z = -2;

Step-by-step explanation: ONe way of solving a system of equations is using the Gauss-Jordan Elimination.

The method consists in transforming the system into an augmented matrix, which is writing the system in form of a matrix and then into a Row Echelon Form, which satisfies the following conditions:

There is a row of all zeros at the bottom of the matrix;
The first non-zero element of any row is 1, which called leading role;
The leading row of the first row is to the right of the leading role of the previous row;

For this question, the matrix is a Row Echelon Form and is written as:

$\left[\begin{array}{ccc}1&2&2\\0&1&3\\0&0&1\end{array}\right]$ $\left[\begin{array}{ccc}4\\9\\-2\end{array}\right]$

or in system form:

x + 2y + 2z = 4

y + 3z = 9

z = -2

Now, to determine the variables:

z = -2

y + 3(-2) = 9

y = 15

x + 30 - 4 = 4

x = - 22

The solution is (-22,15,-2).

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3 years ago