I have these two equations and i'm asked to "solve the system by graphing"....?

Question

So where do i start and what do i do?

first one is:
3x + y = 6
3x – y = 0

And the second one is:
x – y = 2
3x – 3y = 6


can anyone explain this to me a little....my teacher isn't very helpful!

...and i have no idea how to get them into y=mx+b...or something like that

Answer 1

The standard form for an equation of a straight line is:

y = mx + c

where

m is the gradient (how steep the line is)
c is the y-intercept (where it cuts the y-axis)


Given 3x + y = 6

Rearrange it to look like the standard form:

3x + y = 6

Substract 3x from both sides,

3x - (3x) + y = 6 - (3x)

y = - 3x + 6


Similarly, given 3x – y = 0

- y = - 3x

Change the signs on both sides (i.e. multiply by -1 on both sides)

y = 3x


To plot the graphs of

y = - 3x + 6

let x = 0, you will get y = -3(0) + 6 = 6
let x = 1, you will get y = -3(1) + 6 = 5

So, plot the points (0, 6) and (1, 5) and draw a straight line across the two points.

Similarly,

y = 3x

let x = 0, y = 3(0) = 0
let x = 1, y = 3(1) = 3

So, plot the points (0, 0) and (1, 3) and draw a straight line across the two points.

Where the two line intersect, find the corresponding x and y values.

The point of intersection would be
x = 1 and y = 3


Try the same method for your second set of equations.

Answer 2

The previous question presumes you know how to create graphs.

To create graphs used checked paper (paper with intercepting lines forming squares). Change the equations to the y=ax+b by keeping all y elements to the left of the equals sign and the x and number elements to the right. Remember the sign changes when an element changes sides (eg 3x becomes -3x when it changes side).

Then you assign an integer value to x (start from 0 and up) and find the value of y. Write down the values in pairs (x, y). Do the same for the second equation.

Next draw a cross on the graph paper by drawing a horizontal line from end to end and a vertical line in the middle of the horizontal on from end to end. The point where they intercept is the (0,0) point. Starting from there make little marks on the lines (it helps if they're on the cross of the paper and assign numbers to them counting from 1 (2,3,4,5,6...). Use the horizontal line as the x-axis and the vertical as the y-axis.

Using the paired values you identified before plot the points that correspond to the values. For example for value (2,3) you'd plot a point above the x-axis 2 and right to the y-axis 3.

Plot all value pairs this way. Draw a line across all those points.

Do the same for the second equation. Now because your equations have been specially picked the point where the two lines intercept is most likely an integer number. This is the solution to the system of equations.

Answer 3

Since you have been given 2 sets of equations, you will get two solutions. Consider the first two equations.
3x+y = 6 ...........(i)
3x - y = 0...........(ii)
Solve y in the eqn.(i) in terms of x as follows:
y = 6 - 3x
Take at least three combinations of values that satisfy the above eqn. For example, take (0,6),(2.0) and (3, -3) and draw a straight line graph.
Again solve y in (ii) in terms of x as follows:
y = 3x
Take at least three combinations of values that satisfy the above eqn. such as (0,3), (1,3) and (-1,-3).
draw another graph which is parallel to x- axis. Find the point of intersection. That point is the values of x and y.
Regarding the last two setsof equations, it is found that the twoare the same. because x - y =2 .......(i)
and 3x - 3y = 6
=> 3(x - y) =3.2
=> x - y = 2 ..........(ii)
Since the two are the same, you will get only one straight line which means there are infinite solutions to the equation.
I hope it helps.

Answer 4

By graphing means you have to plot the graph my dear.
Where the two graphs intercept is your answer for values of x and y.

y= mx +c

okay let take example 3x + y = 6.
All you need to do is move the y to one side eg.

3x + y = 6
y = -3x + 6

There, your answer. so m = -3 and c = 6
For info m is the gradient of your graph and +c is your y intercept (this is point where you graph cuts the y-axis, also known as when x=0)

Got it ?
Drop me a mail and I'll try explain / teach you.

Cheers.

Answer 5

To draw the graph and to locate the point of intersection of the two lines, try with basic figures like 0, +or -1,+ or - 2 ... as the values of x and by putting them in the given equations we get the corresponding values of y. By noting those points on the graph and joining them we get the lines. Once the lines are drawn it is easy to locate the point of intersection.
Herein the given equations, it is easy to find the point of intersection just by inspection or by trial and error method.
ie. putting, x=1, in the first equation of the first set of equations, we get y=3, and the same value in the second equation also. this implies that point of intersection is (1,3). Therefore just by giving three values to x , ie. 1,2,3 we can draw the two lines and can easily find the point of intersection of the two lines and simply note down the position of it as solution.
The same method can be applied to the second set of equations to find the solution. But, here it has infinite solutions because the given lines are collinear.

Answer 6

first:
take the first and add the second to it.
you will have 6x=6 => x=1
after that you replace x with 1 in on of the them and have y=3 or so.
the second:
mulitiply the firs by 3 then extract the second from it.
you will have 3x - 3y =6
and the second is the same 3x - 3y =6 =>
the system has infinetly many answears.
A system to be solvable it has to have so many(diiferent) equations how many the unknowns are.

Answer 7

Sketch the two graphs (first place them into y=mx + c mode), and then find where they intercept. That point is the "solution."

Answer 8

3x + y = 6
y = -3x + 6

3x - y = 0
-y = -3x
y = 3x

Go to the site below and type in

-3x + 6, press enter
3x, press enter

ANS : (1,3)

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x - y = 2
-y = -x + 2
y = x - 2

type in x - 2

3x - 3y = 6
-3y = -3x + 6
y = x - 2

type in x - 2

You will get infinite solutions

For a graph, go to http://www.calculator.com/calcs/GCalc.html