How do I "Solve Systems by Graphing" in Algebra?

Question

Each problem has two equations. Here are my problems (There is 6 of them)

#1.
x-y = 1
4x+3y= 11

#2.
2x-y= -11
-x+2y =13

#3.
y= -2x + 4
2x+y= 4

#4.
y = -6
2x + 2y = -12

#5.
x + y = 4
y = -x + 1

#6.
2x+3y= -1
4x+6y = -2

How do you solve these and what would the answers be? My teacher said if the lines cross the answer is (x,y). If the lines are parallel there is No Solution. And, if the lines are parallel and overlap, there are infinite solutions. So if someone can easily explain to me how to do this, I will appreciate it. Thanks.

Answer 1

if you want to solve by graph you need to solve fory in term of x first ( i mean like number 1. x-y=1 you add both side by y and subtract by 1 so y=x-1 but remember the coefficient in front of y need!!!!! to be 1)
then graph two lines together.
then hit 2nd trace you will see the word"intercept" hit on it. then use the arrow to move the point go to closer to the point they cross. then hit enter three times they will give you x, y
if the parallel( they have the same slope but different y intercept) they have no solution because two lines never!!!! hit each other( cross).
if they have the same slope and the same y - intercept they will give you infinity solutions. come back to your problem now
#1 solution is x=2, y=1
#2 solution is x=-3,y=5
#3 is so kool because in the first we have y=-2x+4
second we have 2x+y=4
the first they all ready solve y in term of x for you
you need to solve the second . subtract both sides by 2x
that will become y=-2x+4 now compare the first equation and second equation you will see two lines are the same( slope =-2, y intercept is 4)
solution is infinity solutions
#4 solution is x=0, y=-6
#5 no solution
because the first equation if you solve for y
subtract x both side you have
y=-x+4 you see they have the same slope m=-1
y=-x+1 but different y intercept 1st have 4, 2nd have 1
#6 1st solve for y =-2x/3 -1/3
2nd solve for y=-2x/3 -1/3
again theyhave the same slope m=-2/3 and y intercept -1/3
solution will be infinity
good luck i hope you understand it

Answer 2

You need to change each of the equations from Standard (Ax+By=C) to Y=Mx+B. You then graph the Y-intercept (that's the B). From the Y-intercept, you graph the slope (that's the Mx part)

Ex. 1. x-y=1 --> y = -x+1 (you'd graph 1 as the Y-intercept then go up one box and to the left.)
4x-4x+3y=11-4x
equals:
3\3y=-4x+11\3
equals:

Y= -4/3x + 11/3 (you’d then find the point for (11,3) and then find the slope)

There are many other ways to figure out the points on a graph. Like Subsitution. Again, using #1 as the example:

You have Y= -x+1.

You then subsitute this equation and put it in as Y in the other equation:

4x+3(-x+1)=11

You can then solve the equation from there by distrubuation. You'd then combine like terms and solve the equation like a 2-step problem. You'd then put the solution to X in x-y=1 and find out what Y is.

There is also another way. And here it is:

You first have to line up the equations:

x-y=1
4x+3y=11

You'd then multpy the top equation by 3.

3x-3y=3
4x+3y=11

You'd subtract the -3y and 3y to cancel them out. You add the rest to get:

7x=14

Divide by 7 to get x=2. Then put 2 back in the equation as X to figure out Y.

For a short cut: to find out if 2 lines are parallel, you can look at the slopes. If the slopes are the same, it's parallel. If both of the equations are the same then it's the same line.

Answer 3

You just graph each equation, and where they cross is the solution.

You want to get each equation to the form y= mx + b, as this is really easy to graph it in. m = the slope, and b = the y-intercept (where it crosses the y-axis).

#1. x - y = 1
x - 1 = y
y = x - 1

4x + 3y = 11
3y = -4x + 11
y = (-4/3)x + (11/3)

Those are you two equations:
y = x - 1
y = (-4/3)x + (11/3)

Either use a graphing calculator, or graph them on graph paper. Remember slope = rise / run. So plot your y - intercept, then go up the rise and across the run to another point. Join the points and extend to create the line. (If the slope is positive, you line should look like "/", if the slope is negative, the line looks like "\".)

use a ruler so you know exactly where they intersect.

The answer is (2,1). Do the same with the other numbers.

Answer 4

ok - so you are trying to figure out for each pair of lines, which of the possible things are true:
1. do the lines cross?
2. are they parallel?
3. are they the same line.
What you need to do is to graph both lines - either on graph paper or on your calculator. For example, #4: y = -6 is a horizontal line where y is -6 (6 units below the x-axis) for all values of x. Now, graph 2x + 2y = -12 by making at least two (ideally three) ordered pairs of (x,y). Pick any value of x, figure out what y is. Graph the line and see where it hits the other one. If you use a graphing calculator, you can use the INTERSECT option (which is on the CALC button at the top) to find the intersection of the two lines. Have your teacher show you that.

Answer 5

Before you graph anything you have to put y by its self
so in no. 1
-y=1-x
y= -1+x
3y=11-4x
y=(11/3)-(4/3)x

then type the equations on the y=
then 2nd...calc
then intersect
so the answer to no. 1 will be (2,1)
not sure though just check it again

do just need the answers because if you just need the answer then just do the elimination(i like this better than graphing)

Answer 6

1)solve one of the equations for y.
2)substitue y in the other equation.
3)solve the new equation for x
4)sub x into one of the origianl equations.
5)solve equation in step 4 for y.

I'll do the first one
Step 1
x-y=1
-y=1-x
y=-1+x
step 2
4x+2y=11
4x+2(-1+x)=11
you can solve for x now!!!
step 3
4x-2+2x=11
6x=13
x=13/6
step 4
x-y=1 (orignial equation)
13/6-y=1
Step 5
-y=6/6-13/6
y=7/6

Answer 7

first of all you need to set of of the equations to equal "y". For example #1 y=-x-1, y= -4/3x +11/3....then graph....or maybe use a graphing calculater to check.

Answer 8

You need to graph each one. Then wherever they cross, that is your answer for x,y. If they don't there is no solution... Unfortunately, I can't graph for you on this.