Question

1. $3y = 15 - 5x$

standard form
x-intercept
$ax = c$
y-intercept
$by = c$
intercept = ( )
intercept = ( )

Explanation:

Step1: Convert to Standard Form

The standard form of a linear equation is \(Ax + By = C\). Starting with \(3y = 15 - 5x\), we add \(5x\) to both sides:
\(5x + 3y = 15\)

Step2: Find x-intercept (\(Ax = C\) form)

The x-intercept occurs when \(y = 0\). Substitute \(y = 0\) into the original equation \(3y = 15 - 5x\):
\(3(0) = 15 - 5x\)
\(0 = 15 - 5x\)
Add \(5x\) to both sides: \(5x = 15\)
So the x-intercept equation in \(Ax = C\) form is \(5x = 15\), and the x-intercept point is found by solving for \(x\): \(x = \frac{15}{5} = 3\), so the x-intercept is \((3, 0)\).

Step3: Find y-intercept (\(By = C\) form)

The y-intercept occurs when \(x = 0\). Substitute \(x = 0\) into the original equation \(3y = 15 - 5x\):
\(3y = 15 - 5(0)\)
\(3y = 15\)
So the y-intercept equation in \(By = C\) form is \(3y = 15\), and the y-intercept point is found by solving for \(y\): \(y = \frac{15}{3} = 5\), so the y-intercept is \((0, 5)\).

Answer:

Standard Form: \(5x + 3y = 15\)
x-intercept (equation: \(5x = 15\), point: \((3, 0)\))
y-intercept (equation: \(3y = 15\), point: \((0, 5)\))