To determine what kind of conversion was performed on the equation, we need to understand the forms of linear equations:
1. Standard Form: This form of a linear equation is generally written as [tex]\( Ax + By = C \)[/tex], where [tex]\( A \)[/tex], [tex]\( B \)[/tex], and [tex]\( C \)[/tex] are integers, and [tex]\( A \)[/tex] is non-negative.
2. Slope-Intercept Form: This form is written as [tex]\( y = mx + b \)[/tex], where [tex]\( m \)[/tex] is the slope of the line and [tex]\( b \)[/tex] is the y-intercept.
3. Point-Slope Form: This form is written as [tex]\( y - y_1 = m(x - x_1) \)[/tex], where [tex]\( (x_1, y_1) \)[/tex] is a specific point on the line and [tex]\( m \)[/tex] is the slope.
Starting with the original equation:
[tex]\[ -5x + y = -10 \][/tex]
This equation is in standard form ([tex]\( Ax + By = C \)[/tex]).
The student rewrote it as:
[tex]\[ y = 5x - 10 \][/tex]
This equation is in slope-intercept form ([tex]\( y = mx + b \)[/tex]).
To convert from the standard form to the slope-intercept form, we need to solve the standard form equation for [tex]\( y \)[/tex]:
[tex]\[ -5x + y = -10 \][/tex]
Add [tex]\( 5x \)[/tex] to both sides to isolate [tex]\( y \)[/tex]:
[tex]\[ y = 5x - 10 \][/tex]
Thus, the conversion performed here is from standard form to slope-intercept form.
Therefore, the correct answer is:
- A conversion from standard form to slope-intercept form.
