# Thread: solutions of systems of linear equation

1. ## solutions of systems of linear equation

i just want to know the graphical solution and the system of equation of these given numbers: x-2y=5 and 5x-10y=25

also this numbers: 5x-y=6 and x+y=-4

and also this: 2x-y=5 and x+y=4

thank you..
can anyone answer this right now?
i'm in a hurry..

2. Originally Posted by beauty13
i just want to know the graphical solution and the system of equation of these given numbers: x-2y=5 and 5x-10y=25

also this numbers: 5x-y=6 and x+y=-4

and also this: 2x-y=5 and x+y=4

thank you..
can anyone answer this right now?
i'm in a hurry..
A solution to a system is when the lines cross, which means that the x and y co-ordinates must be equal. In other words, the equations must be equal.
5x-10y=25
$5x-10y=25$ solve for x

$5x=10y+25$

$x=2y+5$

$x-2y=5$ Substitute the value of x into this equation

$\left(2y+5\right)-2y=5$ addition requires no paranthesis...

$2y+5-2y=5$ subtract 5 from both sides

$2y-2y=0$ solve

$0=0$ When a solution comes out in the form of 0=0 than the two lines that were tested are the same line.
Can you do the other ones now or do you need more help?

3. Originally Posted by beauty13
i just want to know the graphical solution and the system of equation of these given numbers: x-2y=5 and 5x-10y=25

also this numbers: 5x-y=6 and x+y=-4

and also this: 2x-y=5 and x+y=4...
Hello,

there are a lot of different methods to solve systems of linear equation. One of these methods (I know it under the name substitution method) was demonstrated by Quick.

Now I'll show you another way with your 2nd problem:

Solve both equations for y:

$5x-y=6 \Longrightarrow y_1=5x-6$ and

$x+y=-4 \Longrightarrow y_2=-x-4$

As Quick pointed out: At the interception the x- and the y-values must be the same in both equations:

$y_1=y_2\rightarrow \ 5x-6=-x-4$ Solve this equation for x and you'll get $x = \frac{1}{3}$
Plug in this value in one of the two equations (it doesn't matter which one you take!) and you'll get $y=-\frac{13}{3}$

The next problem is on your own. (3,1)

Greetings

EB