To determine which system of equations (two or more equations
containing the same set of variables [symbols used to represent
unknown numbers]) represents the graph of the two lines, the student
could write each equation in slope-intercept form (y = mx + b, where m
is the slope and b is the y-intercept) and then convert each equation to
standard form (ax + by = c where a, b, and c are integers).
To find the equation for line a, the student could have used the first
two sets of values in the table and applied the slope formula,
, resulting in . Next, the student
could have substituted one of the ordered pairs from the table, (4, –9),
and the slope, m = –6, into y = mx + b and solved for b, resulting in
–9 = –6(4) + b → –9 = –24 + b → 15 = b. Since b = 15 and m = –6, the
equation for line a in slope-intercept form is y = –6x + 15. To convert
the equation from slope-intercept form to standard form, the student
could have added 6x to both sides of the equation, resulting in
6x + y = 15.
To find the equation for line b, the student could have used the first
two sets of values in the table and applied the slope formula,
, resulting in . Next, the student
could have substituted one of the ordered pairs from the table,
(1, –12), and the slope, m = –3, into y = mx + b and solved for b,
resulting in –12 = –3(1) + b → –12 = –3 + b → –9 = b. Since b = –9 and
m = –3, the equation for line b in slope-intercept form is y = –3x – 9. To
convert the equation from slope-intercept form to standard form, the
student could have added 3x to both sides of the equation, resulting in
3x + y = –9.
This is an efficient way to solve the problem; however, other methods
could be used to solve the problem correctly.
The student likely identified the slopes as positive instead of negative
and multiplied the y-term and constant by the coefficient of the x-term
in each equation, resulting in 6y = x + 90 for line a and 3y = x – 27 for
line b. The student likely then converted the equations from slope-
intercept form to standard form by subtracting the y-term and the
constant value from both sides of the equations, obtaining x – 6y = –90
for line a and x – 3y = 27 for line b. The student needs to focus on
understanding how to write a linear function in standard form when
given a table.