For
|a|=b, solve for
a=b and a=-b
so
|a-4|=3a-6
solve
a-4=3a-6 and a-4=-(3a-6)
first one
a-4=3a-6
minus a from both sides
-4=2a-6
add 6 to both sides
2=2a
divide by 2
1=a
other
a-4=-(3a-6)
a-4=-3a+6
add 3a both sides
4a-4=6
add 4 to both sides
4a=10
divide by 4
a=10/4
a=5/2
a=5/2 and 1
The product of the matrices is an identity matrix. Therefore, X and A are inverse of each other.
The matrices are given as:
![X = \left[\begin{array}{cc}-1&-3\\4&2\end{array}\right]](https://tex.z-dn.net/?f=X%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-1%26-3%5C%5C4%262%5Cend%7Barray%7D%5Cright%5D)
![A = \left[\begin{array}{cc}\frac 15&\frac 3{10}\\-\frac 25 &-\frac 1{10}\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Cfrac%2015%26%5Cfrac%203%7B10%7D%5C%5C-%5Cfrac%2025%20%26-%5Cfrac%201%7B10%7D%5Cend%7Barray%7D%5Cright%5D)
To check the matrices are inverse, we calculate their products.
![A \times X = \left[\begin{array}{cc}\frac 15&\frac 3{10}\\-\frac 25 &-\frac 1{10}\end{array}\right] \times \left[\begin{array}{cc}-1&-3\\4&2\end{array}\right]](https://tex.z-dn.net/?f=A%20%5Ctimes%20X%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Cfrac%2015%26%5Cfrac%203%7B10%7D%5C%5C-%5Cfrac%2025%20%26-%5Cfrac%201%7B10%7D%5Cend%7Barray%7D%5Cright%5D%20%5Ctimes%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-1%26-3%5C%5C4%262%5Cend%7Barray%7D%5Cright%5D)
Multiply the rows of A by the column of X.
This gives
![A \times X = \left[\begin{array}{cc}\frac 15 \times -1 + \frac{3}{10} \times 4&\frac 15 \times -3 + \frac{3}{10} \times 2\\ -\frac 25 \times -1 - \frac{1}{10} \times 4&-\frac 25 \times -3 + -\frac{1}{10} \times 2\end{array}\right]](https://tex.z-dn.net/?f=A%20%5Ctimes%20X%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Cfrac%2015%20%5Ctimes%20-1%20%2B%20%5Cfrac%7B3%7D%7B10%7D%20%5Ctimes%204%26%5Cfrac%2015%20%5Ctimes%20-3%20%2B%20%5Cfrac%7B3%7D%7B10%7D%20%5Ctimes%202%5C%5C%20-%5Cfrac%2025%20%5Ctimes%20-1%20-%20%5Cfrac%7B1%7D%7B10%7D%20%5Ctimes%204%26-%5Cfrac%2025%20%5Ctimes%20-3%20%2B%20-%5Cfrac%7B1%7D%7B10%7D%20%5Ctimes%202%5Cend%7Barray%7D%5Cright%5D)
![A \times X = \left[\begin{array}{cc}1&0\\0&1\end{array}\right]](https://tex.z-dn.net/?f=A%20%5Ctimes%20X%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D)
The product of A and X is an identity matrix.
This means that; both matrix are inverse of each other
Read more about matrix at:
brainly.com/question/4017205
Answer:
4.002 x 10^3
Step-by-step explanation:
I would help but I'm not sure what you are asking for us to do
A system of equations with infinitely many solutions is a system where the two equations are identical. The lines coincide. Anything that is equal to

will work. You could try multiply the entire equation by some number, or moving terms around, or adding terms to both sides, or any combination of operations that you apply to the entire equation.
You could multiply the whole thing by 4.5 to get

. If you want, you could mix things up and write it in slope-intercept form:

. The point is, anything that is equivalent to the original equation will give infinitely many solutions x and y. You can test this by plugging in values x and y and seeing the answers!
The attached graph shows that four different equations are really the same.