Question

regularity write a linear equation in standard form that satisfies the given set of conditions. slope of 0.6, passes through (1, 1)

Explanation:

Step1: Recall point - slope form

The point - slope form of a linear equation is $y - y_1=m(x - x_1)$, where $m$ is the slope and $(x_1,y_1)$ is a point on the line. Given that $m = 0.6=\frac{3}{5}$ and the point $(x_1,y_1)=(1,1)$. Substitute these values into the point - slope form:
$y - 1=\frac{3}{5}(x - 1)$

Step2: Convert to standard form

The standard form of a linear equation is $Ax+By = C$, where $A$, $B$, and $C$ are integers and $A\geq0$.
First, multiply both sides of $y - 1=\frac{3}{5}(x - 1)$ by 5 to eliminate the fraction:
$5(y - 1)=3(x - 1)$
Expand both sides:
$5y-5 = 3x-3$
Rearrange the terms to get the $x$ and $y$ terms on the left - hand side and the constant on the right - hand side:
$- 3x+5y=5 - 3$
Simplify the right - hand side:
$-3x + 5y=2$
Multiply both sides by - 1 to make the coefficient of $x$ positive:
$3x-5y=-2$

Answer:

The linear equation in standard form is $3x - 5y=-2$