Question

How many solutions does the system below have? y = x+1 2y - x = 2

Answer 1

Only one solution

$y = x + 1 \\ \implies \: x - y + 1 = 0..(1) \\\implies \: a_1x + b_1y+ c_1 = 0 \\ a_1 = 1, \: \:b_1 = - 1, \: \: c_1 = 1\\ 2y - x = 2 \\ \implies x - 2y + 2 = 0 \\ \implies \: a_2x + b_2y + c_2 = 0 \\ a_2 = 1, \: \:b_2 = - 2, \: \: c_2 = 2 \\ \\ \frac{a_1}{a_2} = \frac{1}{1} = 1 \\ \\ \frac{b_1}{b_2} = \frac{ - 1}{ - 2} = \frac{ 1}{ 2} \\ \\ \because \: \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \\ \therefore \: given \: system \: of \: linear \: equations \: \\ has \: \bold\red{\boxed{only \: one}} \: solution.$

Answer 2

Answer:

1

Step-by-step explanation:

There is only 2 equations, and 2 unknowns shown on this question. Which means that there could only be one point of intersection.