Solving Linear Systems of Equations
Example
A system of two linear equations has the solution (-2, 2).
The slope of one line is one-quarter the slope of the other line. If the
equation of one line is y = 2x + 6, what are the two possible equations of
the other line?
Solution
| We are given the equation y = 2x + 6. Let’s
call the slope of this line m1. |
m1 = 2 |
| • Possibility 1 Find the equation of the line through (-2, 2) with slope
one-quarter of m . |
| Find the slope, m2, of the new line.
We are given that this second line
passes through the point (-2, 2).
We can use the point-slope formula to
determine the equation of the line. |
y - y1 = m(x - x1) |
Substitute 2 for y1,
 for m, and
-2 for x1.
Solve for y. |
 |
Let’s check to be sure the equation satisfies the given conditions.
| Condition
• The slope of one line is one-
quarter the slope of the other
line. |
Check
• The slope of the given line,
y = 2x + 6, is 2.
The slope of the new line,
, which is onequarter
of 2. |
| • The new line passes through
the point (-2, 2). |
• Substitute -2 for x and 2 for y in
the new equation and simplify.
 |
| • Possibility 2 Find the equation of the line through (-2, 2) with slope 4
times m1, the slope of the given line.
In this case, the slope of the given line, y = 2x + 6, is one-quarter that
of the new line. |
| Let’s call the slope of this new line m3.
Substitute 2 for m1.
Multiply both sides by 4. |
 |
| Now, we know the slope of the new line is 8. Also, the line passes
through (-2, 2). |
| To find the equation of the new line, we use
the point-slope formula as in Possibility 1.
Substitute 2 for y1, 8 for m, -2 for x1.
Solve for y. |
y - y1
y - 2
y |
= m(x - x1)
= 8(x - (-2))
= 8x + 18 |
We leave it to you to check that this equation satisfies the given
conditions.
|