The given matrix equation is,
.
Multiplying the matrices with the scalars, the given equation becomes,
![\left[\begin{array}{cc}1.5x&9\\12&6\end{array}\right] +\left[\begin{array}{cc}y&4y\\3y&2y\end{array}\right] =\left[\begin{array}{cc}z&z\\6z&2\end{array}\right] \\](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1.5x%269%5C%5C12%266%5Cend%7Barray%7D%5Cright%5D%20%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dy%264y%5C%5C3y%262y%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dz%26z%5C%5C6z%262%5Cend%7Barray%7D%5Cright%5D%20%20%5C%5C%20%20)
Adding the matrices,
![\left[\begin{array}{cc}1.5x+y&9+4y\\12+3y&6+2y\end{array}\right] =\left[\begin{array}{cc}z&z\\6z&2\end{array}\right] \\](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1.5x%2By%269%2B4y%5C%5C12%2B3y%266%2B2y%5Cend%7Barray%7D%5Cright%5D%20%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dz%26z%5C%5C6z%262%5Cend%7Barray%7D%5Cright%5D%20%20%5C%5C%20)
Matrix equality gives,

Solving the equations together,

We can see that the equations are not consistent.
There is no solution.
Answer:
A= X,B
B= L,G
C= M,W
The total number of combinations is 12
<span>Step 1: 0.46 = 46⁄100</span>
<span>Step 2: Simplify 46⁄100 = 23⁄50</span><span> hope this helped u</span>
The answer to this equation is
x= 1, -3
Rewrite the given quadratic equation in standard form: Kx 2 + 2x - 1 = 0
Discriminant = 4 - 4(K)(-1) = 4 + 4K
For the equation to have two real solutions, the discriminant has to be positive. Hence we need to solve the inequality 4 + 4K > 0.
The solution set to the above inequality is given by: K > -1 for which the given equation has two real solutions.