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Solve the system by addition.
–5x + 2y = 10
x – 3y = 11
--------------------------------------
Solve the system by graphing.
2x + y = 4
x + y = 3

thanks

2007-02-09 18:13:15 · 4 answers · asked by dee free 2 in Science & Mathematics Mathematics

4 answers

Solve the system by addition.
–5x + 2y = 10
x – 3y = 11

Simple:
Multiply the second line by 5:
-5x + 2y = 10
5x - 15y = 55

Now add the first and second line:
-13y = 65

Solve for y, dividing both sides by -13:
y = -5

Replace y on any of the two lines, like the first line, for example:
-5x + (2*-5) = 10
-5x -10 = 10

Add 10 to both sides:
-5x = 20
x = -4

Solution:
x = -4 and y = -5

------------------------------
Solve the system by graphing.
2x + y = 4
x + y = 3

Simply change the equations around, leaving y on one side and x and the number on the other side, like this:
y = -2x + 4
y = -x + 3

Now, create a table with values for x and y and do the graphing (you should have learned this already, otherwise they would not have asked for the graphic at all).

The solution is where both curves intersect (or cross) each other.

The solution, if the graphic is done correctly, should be:
x = 1 and y = 2

2007-02-09 18:25:54 · answer #1 · answered by F B 3 · 0 0

multiply the second by 5
–5x + 2y = 10
5x –15y = 55
----------------------------
-13y=65

y=-5

x=11-3*5
=11-15

x=-4

2)
draw the graphs of the lines

y=-2x+4 (cuts y at 4 , cuts x at 2)

y=-x+3 (cuts y at 3, cuts x at 3)

2007-02-10 02:25:29 · answer #2 · answered by iyiogrenci 6 · 0 0

A.
Add five time the lower equation to the top to cancel the x
and solve for y.
Add three times the top equation to two times the bottom equation to cancel the y and solve for x
B.
Convert the equations to y = mx + b form
y = -2x + 4
y = -1x + 3
Graph as two lines, where they intersect is the answer.

2007-02-10 02:23:29 · answer #3 · answered by a simple man 6 · 0 0

[-5x + 5x] + [2y + 5*(-3y)] = 10 + 5*11
-13y = 65
y = -65/13 = -5
x = 11+3y = 11+3(5)= 26

(26,-5)

---------------------
(1,2)

2007-02-10 02:21:49 · answer #4 · answered by Anonymous · 0 0

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