Question

1. $9y + 6 = 3x$ $x + 2 = 3y$

$m = \underline{\quad\quad}$ $m = \underline{\quad\quad}$
$b = \underline{\quad\quad}$ $b = \underline{\quad\quad}$
circle: one, none, infinite

Explanation:

Step1: Rewrite first equation in slope - intercept form

The slope - intercept form of a linear equation is $y = mx + b$, where $m$ is the slope and $b$ is the y - intercept.
For the equation $9y+6 = 3x$, we solve for $y$:
Subtract 6 from both sides: $9y=3x - 6$
Divide both sides by 9: $y=\frac{3x - 6}{9}=\frac{1}{3}x-\frac{6}{9}=\frac{1}{3}x-\frac{2}{3}$
So for the first equation, $m=\frac{1}{3}$ and $b =-\frac{2}{3}$

Step2: Rewrite second equation in slope - intercept form

For the equation $x + 2=3y$, we solve for $y$:
Divide both sides by 3: $y=\frac{1}{3}x+\frac{2}{3}$
So for the second equation, $m=\frac{1}{3}$ and $b=\frac{2}{3}$

Step3: Analyze the number of solutions

Two linear equations $y = m_1x + b_1$ and $y=m_2x + b_2$ have:

• One solution if $m_1

eq m_2$

• No solution if $m_1 = m_2$ and $b_1

eq b_2$

• Infinite solutions if $m_1 = m_2$ and $b_1 = b_2$

Here, $m_1=m_2=\frac{1}{3}$ and $b_1=-\frac{2}{3}
eq b_2=\frac{2}{3}$, so the lines are parallel and there are no solutions.

Answer:

For the first equation: $m=\frac{1}{3}$, $b =-\frac{2}{3}$; For the second equation: $m=\frac{1}{3}$, $b=\frac{2}{3}$; The number of solutions: none