<span>4*((‑(8*x))+5)-(‑(33*x))-26
sorry if this is to hard to understand
</span>
The solution to the system of equations x + 2y = 1 and -3x-2y = 5 is:
x = -3, y = 2
The given system of equations:
x + 2y = 1............(1)
-3x - 2y = 5..........(2)
This can be written in matrix form as shown:
![\left[\begin{array}{ccc}1&2\\-3&-2\end{array}\right] \left[\begin{array}{ccc}x\\y\end{array}\right] = \left[\begin{array}{ccc}1\\5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%5C%5C-3%26-2%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%5C%5C5%5Cend%7Barray%7D%5Cright%5D)
Find the determinant of ![\left[\begin{array}{ccc}1&2\\-3&-2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%5C%5C-3%26-2%5Cend%7Barray%7D%5Cright%5D)

![\triangle_x = \left[\begin{array}{ccc}1&2\\5&-2\end{array}\right]\\\triangle_x = 1(-2)-2(5)\\\triangle_x = -2-10\\\triangle_x =-12](https://tex.z-dn.net/?f=%5Ctriangle_x%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%5C%5C5%26-2%5Cend%7Barray%7D%5Cright%5D%5C%5C%5Ctriangle_x%20%3D%201%28-2%29-2%285%29%5C%5C%5Ctriangle_x%20%3D%20-2-10%5C%5C%5Ctriangle_x%20%3D-12)
![\triangle_y = \left[\begin{array}{ccc}1&1\\-3&5\end{array}\right]\\\triangle_y = 1(5)-1(-3)\\\triangle_y = 5 + 3\\\triangle_y =8](https://tex.z-dn.net/?f=%5Ctriangle_y%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%5C%5C-3%265%5Cend%7Barray%7D%5Cright%5D%5C%5C%5Ctriangle_y%20%3D%201%285%29-1%28-3%29%5C%5C%5Ctriangle_y%20%3D%205%20%2B%203%5C%5C%5Ctriangle_y%20%3D8)


The solution to the system of equations x + 2y = 1 and -3x-2y = 5 is:
x = -3, y = 2
Learn more here: brainly.com/question/4428059
Answer:
y=63
Step-by-step explanation:
PEMDAS states that the value in the parenthesis should be evaluated first:
4(5*6)
4(30)
After just multiply the value in the parenthesis by the number outside of the parenthesis:
120
There's two ways to do this.
The first way:
Work out the amount of sugar needed to make one cake, and then multiply that by 7.
30 grams ÷ 4 cakes = 7.5 grams for 1 cake.
7.5 grams × 7 cakes = 52.5 grams for 7 cakes.
The second way:
Work out the ratio of sugar:cake by doing 7 ÷ 4, and then multiplying that value by 30.
7 cakes ÷ 4 cakes = 1.75
1.75 × 30 grams = 52.5 grams for 7 cakes.
Either way, the answer is 52.5 grams.