Question

determining a number of solutions
determine whether the systems have one solution, no solution, or infinitely many solutions.
$3x - 2y = 1, 6x - 4y = 1$
$3x - 5y = 0, 5x - 3y = 2$
$3x + 2y = 8, 4x + 3y = 1$
$3x - 6y = 3, 2x - 4y = 2$
$5x - 4y = 2, 5x - 9y = 1$
one solution
no solution
infinitely many solutions

Explanation:

For each system, we compare the ratios of coefficients of $x$, $y$, and constants (for linear systems $a_1x+b_1y=c_1$, $a_2x+b_2y=c_2$):

• If $\frac{a_1}{a_2}

eq \frac{b_1}{b_2}$: One solution

• If $\frac{a_1}{a_2} = \frac{b_1}{b_2}

eq \frac{c_1}{c_2}$: No solution

• If $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$: Infinitely many solutions

Step1: Analyze System 1

System: $3x-2y=3$; $6x-4y=1$
Ratios: $\frac{3}{6}=\frac{1}{2}$, $\frac{-2}{-4}=\frac{1}{2}$, $\frac{3}{1}=3$. $\frac{1}{2}=\frac{1}{2}
eq3$

Step2: Analyze System 2

System: $3x-5y=8$; $6x-3y=2$
Ratios: $\frac{3}{6}=\frac{1}{2}$, $\frac{-5}{-3}=\frac{5}{3}$. $\frac{1}{2}
eq\frac{5}{3}$

Step3: Analyze System 3

System: $3x+2y=8$; $4x+3y=1$
Ratios: $\frac{3}{4}$, $\frac{2}{3}$. $\frac{3}{4}
eq\frac{2}{3}$

Step4: Analyze System 4

System: $3x-6y=3$; $2x-4y=2$
Ratios: $\frac{3}{2}$, $\frac{-6}{-4}=\frac{3}{2}$, $\frac{3}{2}$. $\frac{3}{2}=\frac{3}{2}=\frac{3}{2}$

Step5: Analyze System 5

System: $3x-4y=2$; $6x-8y=1$
Ratios: $\frac{3}{6}=\frac{1}{2}$, $\frac{-4}{-8}=\frac{1}{2}$, $\frac{2}{1}=2$. $\frac{1}{2}=\frac{1}{2}
eq2$

Answer:

One Solution:
$3x - 5y = 8; 6x - 3y = 2$
$3x + 2y = 8; 4x + 3y = 1$

No Solution:
$3x - 2y = 3; 6x - 4y = 1$
$3x - 4y = 2; 6x - 8y = 1$

Infinitely Many Solutions:
$3x - 6y = 3; 2x - 4y = 2$