How do you solve this system of linear equations: 3x+y=3 & 4x-2y=16?

Answer 1

first off, isolate Y in 3x+y=3 so: y=-3x+3

then plug it into 4x-2y=16
so: 4x-2*(-3x+3)=16
4x+6x-6=16
10x=10
x=1
now plug 1 into formula
so: y=-3(1)+3
y=0

SO:(1,0)

Answer 2

1) 3x + y = 3
2) 4x - 2y = 16

There are various ways to solve a system like this.

Substitution: solve equation 1 for y and substitute that result into equation 2

y = 3 - 3x
4x - 2(3 - 3x) = 16
4x - 6 + 6x = 16
10x = 22
x = 22/10 = 11/5

Put into equation 1

3(11/5) + y = 3
33/5 + y = 3
y = 3 - 33/5 = 15/5 - 33/5 = - 18/5

x = 11/5, y = -18/5

Check
4x - 2y = 16
4(11/5) - 2(-18/5) = 16
44/5 + 36/5 = 16
80/5 = 16
16 = 16

Addition: multiply equation 1 by 2 and add the equations

6x + 2y = 6
4x - 2y = 16
----------------
10x = 22
x = 11/5

Answer 3

There are several methods. A simple one is combining the equations to eliminate a variable. If you multiply the first equation by 2 you get

6x + 2y = 6; Note that the second equation contains -2y, so add them together to get

10x = 22; x = 22/10;


You could also multiply the first equation by -4/3 to get

-4x - (4/3)y = -4; then when you add this to the secon you get

-(2+4/3)y = 12. y = -36/10

Answer 4

use linear combination to cancel out a variable
3x+y=3 multiply this by 2 to cancel out the -2y in the other equation
6x+2y=6
+4x-2y=16
--------------
10x=22
x=22/10=2.2
plug x back into the equation and solve for y=-3.6

Answer 5

Solve one equation for x in terms of y, then substitute.
3x+y=3
divide by 3
x + y/3 = 1
subtract y/3
x = 1- y/3
NOW substitute in this equation: 4x-2y=16
4(1-y/3) -2y = 16
4 -4y/3 -2y = 16
Multiply by 3
12 -4y -6y = 48
12 -10y = 48
subtract 48
-36 - 10y = 0
add 10y
-36 = 10y
divide by 10
-3.6 = y

now to solve for x, put -3.6 instead of y into the above equation (either one)
3x + (-3.6) = 3
divide by 3
x - 1.2 = 0
add 1.2
x = 1.2

Answer 6

3x+y=3
4x-2y=16
multiply first by 2
and add in second
6x+2y+4x-2y=6+16
10x=22
x=11/5
y=3-3*11/5
=-18/5

Answer 7

Solve for one of the variables in one of the equations and then substitute that into the other equation (for the variable you sovled for), and solve that equation. (x+y=6, 2x+y=12:x=6-y:2(6-y)+y=12). If you have a graphic calculator (pref a TI-83 plus) you could solve both for x and plug them into the functions and find where they intersect, that will be the answer.

Answer 8

First you need to get the Y by itself.

Y = -3x + 3

Y = 2x - 8

Next, you need to graph. Graph equation number 1. You do that by graphing the y-intercept, (3) by putting the first point on the 3rd line up on the y-axis. From there you go down three, forward one, Put a dot. Again, down three, forward one, and put another dot. Now you have three dots. Draw a line.

Now graph equation number 2. Y-intercept (-8) is 8 points below the origin. From there you go up two, forward one, Put a dot. Again, up two, forward one, and put another dot. You have three dots, now draw a line.

There, you've graphed both lines. I'm not sure whether from there you need to put the intercepting coordinates or just leave it like that, but you know what to do.

Hope this helped.

(If you didn't need the graph, this was worth nothing. Sorry.)
(Say what you need.)

Answer 9

2 * 3x+y=3.....becomes 6x+2y=6....add this to 4x-2y=16......this equals....10x=22.....x=11/5........then put x back into either and solve for y......y= -3.6

Answer 10

you ought to instruct them into equations. So, the 1st could be y=2x-2. the 2nd could be y=4x-10 and y=-3x+11. i don't be attentive to the 1st one, or the 0.33 one. i'm hoping I helped a minimum of a splash!