+0

# Can someone solve this?

0
204
4

xy+4x=2880

xy-12y=2160

x and y in both equations is the same

Guest Feb 20, 2017

#4
+19661
+20

xy+4x=2880

xy-12y=2160

x and y in both equations is the same

$$\begin{array}{|lrcll|} \hline (1) & xy+4x &=& 2880 \\ (2) & xy-12y &=& 2160 \\ \hline \\ (1)-(2): & xy+4x -(xy-12y)&=& 2880-2160 \\ & xy+4x -xy+12y &=& 720 \\ & 4x+12y &=& 720 \quad & | \quad : 4 \\ & x+3y &=& 180 \\ (3) & x &=& 180-3y \\ \hline \end{array}$$

(3) in (2):

$$\begin{array}{|rcll|} \hline (2) & xy-12y &=& 2160 \\ & y\cdot (x-12) &=& 2160 \quad & | \quad x = 180-3y \\ & y\cdot (180-3y-12) &=& 2160 \\ & 180y-3y^2-12y &=& 2160 \\ & -3y^2+168y &=& 2160 \quad & | \quad : (-3) \\ & y^2-56y &=&-720 \\ & y^2-56y +720 &=& 0 \\ & (y-36)(y-20) &=& 0 \\\\ & y_1 = 36 &\text{or}& \quad y_2 = 20 \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline (3) & x &=& 180-3y \\ & x_1 &=& 180-3\cdot 36 \\ & x_1 &=& 72 \\\\ & x_2 &=& 180-3\cdot 20 \\ & x_2 &=& 120 \\ \hline \end{array}$$

Solutions: (72,36) and (120,20)

heureka  Feb 21, 2017
#1
0

xy+4x=2880

xy-12y=2160

x = 72 and y = 36

x = 120 and y = 20

Guest Feb 20, 2017
#3
+87334
0

xy + 4x= 2880    →  x (y + 4)  = 2880 →  x = 2880/ (y + 4)    (1)

xy - 12y= 2160     (2)

Sub (1)  into (2)

[2880/ (y + 4)] y - 12y  = 2160    multiply through by ( y + 4)

2880y - 12y( y + 4) = 2160( y + 4)    simplify

2880y - 12y^2 - 48y = 2160y + 8640

12y^2 - 672 y  + 8640  = 0

y^2 - 56y + 720  = 0     factor

( y - 36) (y - 20)  = 0

Set each factor to 0 and y = 36  or y  = 20

When y  = 36,  x = 2880/[36 + 4]  = 2880/40 =  72

When y  = 20,  x = 2880/[20 + 4]  = 2880/24 =  120

Solutions    (72, 36)  and (120, 20 )

CPhill  Feb 20, 2017
#4
+19661
+20

xy+4x=2880

xy-12y=2160

x and y in both equations is the same

$$\begin{array}{|lrcll|} \hline (1) & xy+4x &=& 2880 \\ (2) & xy-12y &=& 2160 \\ \hline \\ (1)-(2): & xy+4x -(xy-12y)&=& 2880-2160 \\ & xy+4x -xy+12y &=& 720 \\ & 4x+12y &=& 720 \quad & | \quad : 4 \\ & x+3y &=& 180 \\ (3) & x &=& 180-3y \\ \hline \end{array}$$

(3) in (2):

$$\begin{array}{|rcll|} \hline (2) & xy-12y &=& 2160 \\ & y\cdot (x-12) &=& 2160 \quad & | \quad x = 180-3y \\ & y\cdot (180-3y-12) &=& 2160 \\ & 180y-3y^2-12y &=& 2160 \\ & -3y^2+168y &=& 2160 \quad & | \quad : (-3) \\ & y^2-56y &=&-720 \\ & y^2-56y +720 &=& 0 \\ & (y-36)(y-20) &=& 0 \\\\ & y_1 = 36 &\text{or}& \quad y_2 = 20 \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline (3) & x &=& 180-3y \\ & x_1 &=& 180-3\cdot 36 \\ & x_1 &=& 72 \\\\ & x_2 &=& 180-3\cdot 20 \\ & x_2 &=& 120 \\ \hline \end{array}$$

Solutions: (72,36) and (120,20)

heureka  Feb 21, 2017