Question

How many solutions does this linear system have? y = 2x – 5 –8x – 4y = –20

Answer 1

The second one would be y=-2x+5, and it would only have one solution.

solution is (2.5, 0)

Answer 2

Answer:

Step-by-step explanation:

Equations $a_1x+b_1y+c_1=0\,\,,\,\,a_2x+b_2y+c_2=0$  have

1.unique solution if $\frac{a_1}{a_2}\neq \frac{b_1}{b_2}$

2. infinite solutions if $\frac{a_1}{a_2}= \frac{b_1}{b_2}=\frac{c_1}{c_2}$

3. no solution if $\frac{a_1}{a_2}= \frac{b_1}{b_2}\neq \frac{c_1}{c_2}$

We can write equations: $y=2x-5\,\,,\,\,-8x-4y=-20$ as follows:

$2x-y-5=0\,\,,\,\,-8x-4y+20=0$

On comparing these equations with standard equations $a_1x+b_1y+c_1=0\,\,,\,\,a_2x+b_2y+c_2=0$ , we get $a_1=2\,,b_1=-1\,,\,c_1=-5\,,a_2=-8\,,b_2=-4\,,c_2=20$

such that

$\frac{a_1}{a_2}=\frac{2}{-8}=\frac{-1}{4}\\\frac{b_1}{b_2}=\frac{-1}{-4}=\frac{1}{4}\\\frac{c_1}{c_2}=\frac{-5}{20}=\frac{-1}{4}$

Here, $\frac{-1}{4}=\frac{a_1}{a_2}\neq \frac{b_1}{b_2}=\frac{1}{4}$

So, according to the facts explained above,

The given linear system has a unique solution i.e only one solution.