Question

Classify each system of equations as having a single solution, no solution, or infinite solutions.

Answer 1

Answer:

Infinite solutions systems.

$y=5-2x\\4x+2y=10$

This system has infinitely many solutions because both equations represents the same line. Notice that if we multiply the first equation by 2, it will results in the same equation than the second one.

$2y=10-4x\\4x+2y=10$

No solution systems.

$x=26-3y\\2x+6y=22$

Let's multiply the first equation by -2 and sum

$-2x=-52+6y\\2x+6y=22$

$6y=-30+6y\\6y-6y=-30\\0=-30$

Notice that the result is false. Therefore, this system has no solution.

$x+2y=3\\4x+8y=15$

Let's multiply the first equation by -4 and sum

$-4x-8y=-12\\4x+8y=15$

$0=3$

As you can see, the system has no solutions.

Single solutions system.

$5x+4y=6\\10x-2y=7$

Let's multiply the first equation by -2 and sum

$-10x-8y=-12\\10x-2y=7$

$-10y=-5\\y=\frac{1}{2}$

Now, we find the other value

$5x+4y=6\\5x+4(\frac{1}{2} )=6\\5x+2=6\\x=\frac{6-2}{5}=\frac{4}{5}$

The solution of the system is (4/5, 1/2). Therefore, the system has one solution.

$3x+4y=17\\-6x=10y-39$

Let's multiply the first equation by 2 and sum

$6x+8y=34\\-6x=10y-39$

$8y=10y-5\\5=10y-8y\\2y=5\\y=\frac{5}{2}$

Then, we find the other value

$3x+4y=17\\3x+4(\frac{5}{2} )=17\\3x+10=17\\3x=17-10\\x=\frac{7}{3}$

So, the system has a solution, which is (7/3 , 5/2).

$x+5y=24\\5x=12-y$

Let's multiply the first equation by -5 and sum

$-5x-25y=-120\\5x=12-y$

$-25y=-108-y\\108=25y-y\\24y=108\\y=\frac{108}{24}=4.5$

Then, we find the other value

$x+5y=24\\x+5(4.5)=24\\x=24-22.5\\x=1.5$

Therefore, the system has one solution, which is (1.5, 4.5).

Answer 2

Answer: y=mx+b solution!

Step-by-step explanation: y = 5 - 2x right?

So that's correct and needs no change.

But 4x + 2y = 10 needs to be changed!

4x + 2y =10

-4x          -4x

2y = -4x + 10

__     __   __

2       2     2

y = -2x + 5