how do you solve this?

Question

x(y+2)+y(4-x)=9
x(y+3)+y(8-x)=14

x^2+y=1-x(1-x)
x(x+1)+2y=5-x(2-x)

Answer 1

Hi,

x(y+2)+y(4-x)=9
x(y+3)+y(8-x)=14

Distribute to eliminate the parentheses.

xy + 2x + 4y - xy = 9
xy + 3x + 8y -xy = 14

Combine like terms.

2x + 4y = 9
3x + 8y =14

Multiply the 1st equation by 3 and the second equation by -2. Then add the equations and x terms will drop out. Solve for y.

3(2x + 4y = 9)
-2(3x + 8y =14)

6x + 12y = 27
-6x - 16y = -28
---------------------
-4y = -1
y = ¼

Plug this answer back into either equation and solve for x.

2x + 4y = 9
2x + 4(¼) = 9
2x + 1 = 9
2x = 8
x = 4

The answer is x = 4 and y = ¼ which is the point (4,¼).

To "check" that answer plug both x and y values into the other equation to verify that they work.

3x + 8y =14
3(4) + 8(¼) = 14
12 + 2 = 14
14 = 14

This shows that the answer (4,¼) worked in both equations, so it is correct.

*******************

x² + y = 1 - x(1-x)
x(x+1)+2y=5-x(2-x)

Distribute to eliminate parentheses.

x² + y = 1 - x + x²
x² + x + 2y = 5 - 2x + x²

Combine like terms.

x + y = 1
3x + 2y = 5

Multiply the first equation by -2 and add the equations together. The y term will drop out and then solve for x.

-2(x + y = 1)
3x + 2y = 5

-2(x + y = 1)
3x + 2y = 5

-2x - 2y = - 2
3x + 2y = 5
------------------
x = 3

Plug this answer back into either equation and solve for y.

x + y = 1
3 + y = 1
y = -2

The answer is x = 3 and y = -2 which is the point (3,-2).

To "check" that answer plug both x and y values into the other equation to verify that they work.

3x + 2y = 5
3(3) + 2(-2) = 5
9 - 4 = 5
5 = 5

This shows that the answer (3,-2) worked in both equations, so it is correct.



I hope that helps!! :-)

Answer 2

x(y+2)+y(4-x)=9 ----> xy+2x +4y -xy = 9 ----> 2x+4y=9 #1
x(y+3)+y(8-x)=14 ---> xy+3x+8y-xy = 14 ---> 3x+8y=14 #2
#1 times 2 = 2x+4y= 9 ---> 4x+8y=18 #3
#3 - #2 : x=4 then replace... 8+4y=9 4y=1 y=0.25

ok I'm a little bit lazy for the second one so the answer is x=3 and y= -2

Answer 3

distribute:

xy + 2x +4y -xy =9
xy + 3x + 8y -xy = 14


so, that's

2x +4y = 9
3x +8y = 14

solve one equation for x, you get

x=(9/2) - 2y

stick that into the second equation:

(27/2) -6y +8y = 14


solve for y.


In both problems the xy terms cancel out.

Answer 4

i take it, that we are solving a system of equations?

x(y+2)+y(4-x)=9
x(y+3)+y(8-x)=14

xy+2x+4y-xy=9
xy+3x+8y-xy=14

(2x+4y=9)X-2
3x+8y=14

-4x-8y=-18
3x+8y=14
-x=-4
x=4
y= 1/4 (.25)

x^2+y=1-x(1-x)
x(x+1)+2y=5-x(2-x)

x^2+y=1-x+x^2
x^2+x+2y=5-2x+x^2

y=1-x
3x+2y=5

3x+2(1-x)=5
x=3
y=-2

Answer 5

These are only simultaneous equations in x and y.

The higher degree and other terms cancel out on expansion.