QUICK! Is it possible to find the solution of two linear equations WITHOUT graphing?

Question

ok, so Ive got these 2 equations, both in y=mx+b form, it says do it on a graphing calculator, but 1) I dont know how to actually use mine, and 2) we arnt officially using them in class.

Can someone help me? Fast?

Answer 1

There are several good ways to solve systems of linear equations, and
the best method to use in any given situation is the one that requires
the least amount of work. However, that will depend on the particular
equations that you're trying to solve. The most popular methods are
inspection, elimination, substitution, intersection, and graphing.

Graphing is usually the hardest, unless you have a graphing calculator
or program handy. Then it can be the easiest, but there is always
room for error in trying to read a point of intersection from a graph.
So graphing is good when you just need an approximate answer, but if
you need a lot of precision, you should probably stick with one of the
other methods.

Inspection is the easiest method, and it's always a good idea to check
to see if you can use it before jumping in with another method. When
can you use inspection? It's easiest when one of the variables has
the same coefficient in both equations, e.g.,

3x + 2y = 12
3x + 5y = 18

To use inspection, you reason this way: "In moving from the first
equation to the second, all that changes is that we're adding 3y. On
the right side, we add 6. So it must be the case that 3y is the same
as 6, which means that y must equal 2."

In fact, all you're really doing here is using elimination, but not
bothering to do the formal addition or subtraction of equations. If
we write the equations in the opposite order,

3x + 5y = 18
3x + 2y = 12

we can subtract the left sides and the right sides to get

(3x+5y) - (3x+2y) = (18) - (12)

3x - 3x + 5y - 2y = 6

3y = 6

which is what we did the last time, but using insight instead of
equations. The nice thing about elimination is that it continues to
work even when inspection fails, which is to say, when all the
coefficients are different:

3x + 2y = 12
6x + 5y = 26

To make elimination work in a situation like this, you need to
multiply one of the equations by a constant factor so that you end up
in a situation you like better, i.e., with matching coefficients. In
this case, we can multiply the first equation by 2 to get

6x + 4y = 24
6x + 5y = 26

Now we can use inspection or elimination, depending on how much we
trust ourselves to do the work in our heads instead of on paper.

That case was sort of like adding 1/2 and 1/4, where one of the
denominators is already the one you're going to use as a common
denominator. But sometimes you get equations like

3x + 2y = 12
7x + 5y = 26

And this is where elimination starts to seem like more trouble than
it's worth, because you have to multiply both equations by different
constants to get matching coefficients:

7(3x + 2y) = 7(12)
3(7x + 5y) = 3(26)

At this point, other methods start to look pretty good. The remaining
methods, substitution and intersection, both use the same first step,
which is to select one of the equations and solve it for one of the
variables in terms of the other. In the example above, we might
choose the first equation and do this:

3x + 2y = 12

2y = 12 ?3x

y = 6 ?(3/2)x

At this point, if we want to use substitution, we go back to the other
equation, locate every occurrence of y, and replace it with the
expression 6 ?(3/2)x:

26 = 7x + 5y

= 7x + 5(6 ?(3/2)x)

Now we're in familiar territory again: one equation, with one
variable. We can solve this to find the value of x, and plug that into
either equation to find the value of y.

In intersection, we reason a little differently. Remembering that
we're dealing with lines, we think: "Assuming the lines aren't
parallel, they have to intersect somewhere, and wherever that happens,
they have to have the same value of y (or of x)." So we solve both
equations for the same variable, and set the expressions equal to each
other:

3x + 2y = 12 becomes y = (12 ?3x)/2

7x + 5y = 26 becomes y = (26 ?7x)/5

and since it must be true that y=y, it must also be true that

(12 ?3x) /2 = (26 ?7x)/5

And again, we're in familiar territory.

Personally, I prefer elimination to substitution and intersection,
because I like to work with integers whenever possible. Of course, if
I'm starting out with coefficients that aren't integers, then that
excuse goes out the window, and I'll usually reach for intersection.
(I prefer intersection to substitution because I think the equations
tend to look a little nicer.)

Answer 2

I know a bunch of ways. Graphing on paper is the first and most basic way; however, it's very inexact. You have to LOOK at the graph to find the answers, and you can never be sure that you measure it precisely enough. (Is it really x = 5, or could it be x = 4.998?)

One of the best methods (especially for linear equations) is the Substitution Method. If both are in the form "y = ...," then that y-value has to be the same in both equations at the point of intersection. Take the right side of each equation, and set them equal to each other. Solve for x. Then "back-substitute;" take your found x-value, and plug it into either equation to find y (or plug it in to both in order to check your work; they should yield the same answer).

I'll list the other methods just for reference; find them in your textbook or online: The Elimination Method is very similar to substitution.

Gauss-Jordan Elimination is a special form of the Elimination Method. It can be used on equations in General Form that are translated into matrices.

Another matrix method writes the system of equations as the product of two matrices. Then use the Inverse Matrix to solve for [ x y ].

Cramer's Rule is a method that uses determinants (another function of matrices), and also requres that the equations are linear and in general form.

P. S. If the directions actually tell you to use the graphing calculator, the best thing to do is learn how to use it! There's probably a purpose in making you go through it.

If it's a TI graphing calculator: Go to Y= (which is either a button on the calculator, or it's in the Graph menu). Type in both equations. Graph the equations. Either use Trace to move the cursor to the point of intersection (and zoom in if needed to get a more accurate measure of the answer), or use the "Intersection" command (which will be in different places; it depends on which model of the calculator you have).

Answer 3

if they are both y=mx+b, set them equal to each other. this will eliminate the y variable. then you can easily solve for x
it is actually pretty easy to do on a graphing calculator too.
if you have a TI 83+ like I do, you can just push the y= button in the top left corner. then simply put both of your equations in there and push the graph button (you may have to change the zoom (hit zoom and then option 9 (zoomstat).

Answer 4

Yes there is a way. it's called solving linear equations by substitution.
1) example:
You change the equation to look like this.
y=4x+5 4x+y=5
y=2x+4 (-)2x+y=4
you take 4x+y=5 minus 2x+y=4
The Y's cancel out.
4x+y=5
(-)2x+y=4
-------------
2x =1
you take 1 divide by 2 witch is 1/2
so X= 1/2

you substitute 1/2 for x
It doesn't matter witch equation you use to substitute 1/2 for.

4x1/2+y=5
2+y=5
-2 -2 coordinates are(1/2,3)
-------------
y=3


I Hope this helps you. it's been awhile since I've done these

Answer 5

Yes, since both equations are equal to y, they are equal to each other

example
y=4x + 5 and y = 2x +7

4x + 5 = 2x + 7
2x = 2
x = 1, y = 4(1)+5 = 9

Answer 6

Solve x in terms of y in the first equation, plug first equation into the second equation (substituting x). Now you have an equation with nothing but y. Solve for y, then use your value of y to solve for x in the first (original) equation.

Answer 7

Isolate x

yb = mx

(yb)/m = x

now put the solution for x into the other equation.

solve and simplify.

You can also solve both quations for b, and put them in a matrix.

Answer 8

Of course it is! Just set them equal to each other and solve for x, plug in x to get y, voila

Answer 9

set them equal to each other and solve for x, then plug x into either of them and get y

Answer 10

If you have y =(m1)x +(b1)
and y =(m2)x +b2

subtract one equation from the other
(y -y) = ( (m1)x +(b1) ) -( (m2)x +b2 )
0 = (m1 -m2)x +(b1 -b2)
0 = (m1 -m2)x +(b1 -b2)
(m1 -m2)x = +(b2 -b1)

x = (b2-b1)/(m1-m2)

substitute this value for x into either of the original equations
y =(m1)x +(b1)
y =(m1)((b2-b1)/(m1-m2) )+(b1)