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3 years ago

Solve the following system of equations for all three variables.

Mathematics

1 answer:

Semenov [28]3 years ago

5 0

Answer:

x=2, z=4, y=-6

Step-by-step explanation:

Add

-9x-y+2z= -4

-6x+y+5z= 2

-9x-6x is -15x

-y+y is 0

2z+5z is 7z

-4+2 is -2

So the equation now looks like this: -15x+7z=-2

Subtract

-6x+y+5z= 2

-4x+y+5z= 6

So the equation will look like this:

-6x+y+5z= 2

-(-4x+y+5z= 6)

-6x+4x is -2x

y-y is 0

5z-5z is 0

2-6 is -4

So the equation now looks like this: -2x=-4

You can find x by dividing -2 on both sides:

x=2

So 2 is x

Now you can substitute x into 2

-15(2)+7z=-2

-30+7z=-2

7z=28

Divide 7 on both sides:

z=4

So now you can substitute z into 4 (For any of the three equation)

-4(2)+y+5(4)=6

-8+y+20=6

20 minus 8 is 12

y+12=6

Subtract 12 on both sides:

y=-6

So the answers are

x=2, z=4, y=-6

Hope this helps!

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Step-by-step explanation:

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Evgen [1.6K]

Answer:

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

3. $\left[\begin{array}{ccc}-3&21&60\\-15&9&-45\\\end{array}\right]$

4. $\left[\begin{array}{ccc}6&-14\\-2&-6\\\end{array}\right]$

Step-by-step explanation:

2. This matrix is easy, as it just requires addition.

$\left[\begin{array}{ccc}1&4\\0&3\\\end{array}\right]$ + $\left[\begin{array}{ccc}0&0\\0&0\\\end{array}\right]$ = $\left[\begin{array}{ccc}1&4\\0&3\\\end{array}\right]$

3. This matrix requires for the matrices to be multiplied first, then added.

$\left[\begin{array}{ccc}9&-15&18\\-27&15&-9\\\end{array}\right]$ + $\left[\begin{array}{ccc}-12&36&42 \\12& -6 & -36 \\\end{array}\right]$ = $\left[\begin{array}{ccc}-3&21&60\\-15&9&-45\\\end{array}\right]$

4. Here we can add the last 2 matrices to find x.

$\left[\begin{array}{ccc}2&-8\\-4&2\\\end{array}\right]$ + $\left[\begin{array}{ccc}4&-6\\2&-8 \\\end{array}\right]$ = $\left[\begin{array}{ccc}6&-14\\-2&-6\\\end{array}\right]$

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