Question

19. Find the value of 4x' + x + xy? - y' +6
for r=-2 and y = 5.​

Answer 1

$\huge\underbrace\mathrm\purple{Answer}$

I'm not sure why I have been given a request to answer this question given there are 100+ answers for this already.

What you have here are two equations:

x+y=5x+y=5 [1]

xy=6xy=6 [2]

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

y=5−xy=5−x [3]

Now, substitute this expression into equation [2]:

x(5−x)=6x(5−x)=6

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

5x−x2=65x−x2=6

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

x2−5x+6=0x2−5x+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(x−2)(x−3)=0

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

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

when x=2x=2 , then y=5−2=3y=5−2=3

when x=3x=3 , then y=5−3=2y=5−3=2

Therefore, there are two solutions to these simultaneous equations, which are: (2,3)(2,3) and (3,2)(3,2)