Question

M=x^3+y^2-6(x-y)-2021

Answer 1

Answer:

Step-by-step explanation:

m=x^3+y^2-6(x-y)-2021

distribute: 6x-6y

now you have m=x^3 + y^2 - 6x - 6y - 2021

Answer 2

What you have here are two equations:

x+y=5  [1]

xy=6  [2]

To solve for  x  and  y , you should rearrange one of the equations for one of the variables  x  or  y . Choose equation [1] to rearrange for  y , because it is easy to do so.

y=5−x  [3]

Now, substitute this expression into equation [2]:

x(5−x)=6  

Now use this equation to solve for  x . Firstly expand the parentheses:

5x−x2=6  

Rearrange the terms so that they are all on one side of the equation:

x2−5x+6=0  

What you have here is a quadratic equation. There are many ways to solve a quadratic equation but I will choose the easiest method for this problem.

Factorise the quadratic expression on the left-hand side of the equation by using the product-sum method (or by decomposition):

(x−2)(x−3)=0  

The null factor law gives  x=2  or  x=3  as the solution.

Now, substitute  x=2  or  x=3  into equation [3] to get the value of y:

when  x=2 , then  y=5−2=3  

when  x=3 , then  y=5−3=2  

Therefore, there are two solutions to these simultaneous equations, which are:  (2,3)  and  (3,2)  ■  By inspection, you should be able to see that two numbers whose sum is 5 and whose product is 6 are 2 and 3. It could be that x=2 and y=3, or that x=3 and y=2.

But, one could also solve algebraically by isolating either the x or y in the equation x+y=5, and substituting into the other equation. In that case, x=5–y, so

(5–y)•y=6

5y–y^2=6. Since this is quadratic, get zero on one side and then factor.

y^2–5y+6=0

(y–3)(y–2)=0

y=3 or y=2 Substituting back into the equation x+y=5,

If y=3, x+3=5 so x=2

If y=2, x+2=5, so y=3