Solve Simultaneous Equations y=2x²+3x-31 & y=21-2x

[ladybutterz]

y=2x²+3x-31
y=21-2x
solve for x and y

 

[Evgeny.Makarov]

Equate the right-hand sides.

 

[ladybutterz]

Equate the right-hand sides.

yea but i really need help don't really understand need to see the calculation to understand it properly

 

[MarkFL]

y=2x²+3x-31
y=21-2x
solve for x and y

I have moved this topic as it is a better fit here in the Pre-Algebra and Algebra sub-forum.

The suggestion given by Evgeny.Makarov is an excellent one, and you should be able to make at least some progress with it. If he does it for you, you will learn less than if you attempt it yourself. :D

 

[CellCoree]

To solve for x and y in these simultaneous equations, we can use the substitution method. We know that y is equal to both 2x²+3x-31 and 21-2x, so we can set these two equations equal to each other:

2x²+3x-31 = 21-2x

Next, we can rearrange this equation to put it in standard form:

2x²+5x-52 = 0

We can then use the quadratic formula to solve for x:

x = (-b ± √(b²-4ac))/2a

Plugging in the values from our equation, we get:

x = (-5 ± √(5²-4(2)(-52)))/2(2)

Simplifying this, we get two possible values for x: x = 4 or x = -6.

To find the corresponding values for y, we can plug these values back into either of the original equations:

For x = 4, y = 21-2(4) = 13
For x = -6, y = 21-2(-6) = 33

Therefore, the solutions for x and y are (4, 13) and (-6, 33).