Download Linear Equations Solutions Intermediate Algebra, Joseph Lee

Linear Equations Solutions
Intermediate Algebra, Joseph Lee
(1-3) Write the equation of the line, in slope-intercept form, that has the given slope and passes through the given
point.
1. m = 3, (0, −2)
Solution: For any equation with slope m and y-intercept (0, b), the equation is given as y = mx + b.
Therefore, the equation of this line is
y = 3x − 2.
2
2. m = − , (0, 4)
3
Solution: For any equation with slope m and y-intercept (0, b), the equation is given as y = mx + b.
Therefore, the equation of this line is
2
y = − x + 4.
3
1
3. m = 1, 0,
2
Solution: For any equation with slope m and y-intercept (0, b), the equation is given as y = mx + b.
Therefore, the equation of this line is
1
y =x+ .
2
(4-5) Identify the slope and y-intercept of the line. Write the equation of the line in slope-intercept form.
4. Note: Passes though (0, −1) and (1, 1).
Solution: The y-intercept of the line is (0, −1). To calculate the slope, observe
m=
rise
2
= = 2.
run
1
Thus, the equation of this line is
y = 2x − 1.
5. Note: Passes though (0, 2) and (4, −1).
Linear Equations Solutions
Intermediate Algebra, Joseph Lee
Solution: The y-intercept of the line is (0, 2). To calculate the slope, observe
m=
rise
−3
=
.
run
4
Thus, the equation of this line is
3
y = − x + 2.
4
(6-7) Write the equation of the line, in slope-intercept form, that passes through the given points.
6. (0, 3), (4, −1)
Solution: The y-intercept of the line is (0, 3). To calculate the slope, observe
m=
y2 − y1
−4
−1 − 3
=
= −1.
=
x2 − x1
4−0
4
Thus, the equation of this line is
y = −x + 3.
7. (0, −4), (5, 6)
Solution: The y-intercept of the line is (0, −4). To calculate the slope, observe
m=
10
y2 − y1
6 − (−4)
=
=
= 2.
x2 − x1
5−0
5
Thus, the equation of this line is
y = 2x − 4.
(8-13) Write the equation of the line, in slope-intercept form, that has the given slope and passes through the given
point.
8. m = 3, (2, 5)
Solution: The equation of a line with slope m that passes through a point (x1 , y1 ) is given by y − y1 =
m(x − x1 ). Thus, the equation of the line is
y − 5 = 3(x − 2)
y − 5 = 3x − 6
y = 3x − 1.
Linear Equations Solutions
Intermediate Algebra, Joseph Lee
1
9. m = − , (−4, 1)
2
Solution: The equation of a line with slope m that passes through a point (x1 , y1 ) is given by y − y1 =
m(x − x1 ). Thus, the equation of the line is
1
y − 1 = − (x − (−4))
2
1
y − 1 = − (x + 4)
2
1
y−1= − x−2
2
1
y = − x − 1.
2
2
10. m = − , (6, −5)
3
Solution: The equation of a line with slope m that passes through a point (x1 , y1 ) is given by y − y1 =
m(x − x1 ). Thus, the equation of the line is
2
y − (−5) = − (x − 6)
3
2
y+5= − x+4
3
2
y = − x−1
3
11. m =
4
, (−10, −8)
5
Solution: The equation of a line with slope m that passes through a point (x1 , y1 ) is given by y − y1 =
m(x − x1 ). Thus, the equation of the line is
y − (−8) =
y+8=
4
(x + 10)
5
y+8=
4
x+8
5
y=
12. m =
1
, (−5, 3)
2
4
(x − (−10))
5
4
x
5
Linear Equations Solutions
Intermediate Algebra, Joseph Lee
Solution: The equation of a line with slope m that passes through a point (x1 , y1 ) is given by y − y1 =
m(x − x1 ). Thus, the equation of the line is
y−3=
1
(x − (−5))
2
y−3=
1
(x + 5)
2
y−3=
1
5
x+
2
2
y=
1
11
x+
2
2
3
13. m = − , (6, −5)
4
Solution: The equation of a line with slope m that passes through a point (x1 , y1 ) is given by y − y1 =
m(x − x1 ). Thus, the equation of the line is
3
y − (−5) = − (x − 6)
4
3
9
y+5= − x+
4
2
1
3
y = − x−
4
2
(14-19) Write the equation of the line that passes through the given points. Write the equation in the following:
(a) slope-intercept form
(b) standard form.
14. (3, 4), (5, 10)
Solution: To write the equation, we must find the slope of this line.
m=
y2 − y1
10 − 4
6
=
= = 3.
x2 − x1
5−3
2
We may choose either point to write the equation.
y − 4 = 3(x − 3)
y − 4 = 3x − 9
(a)
y = 3x − 5
To write the equation in standard form, we must move the x’s to the left side, eliminate any fractions, and
make the coefficient of the x term positive.
y = 3x − 5
−3x + y = −5
(b)
3x − y = 5
Linear Equations Solutions
Intermediate Algebra, Joseph Lee
15. (4, −3), (−6, 2)
Solution: To write the equation, we must find the slope of this line.
y2 − y1
2 − (−3)
5
1
=
=
=− .
x2 − x1
−6 − 4
−10
2
m=
We may choose either point to write the equation.
1
y − (−3) = − (x − 4)
2
1
y+3= − x+2
2
1
y = − x−1
2
(a)
To write the equation in standard form, we must move the x’s to the left side, eliminate any fractions, and
make the coefficient of the x term positive.
1
y = − x−1
2
1
x + y = −1
2
(b)
x + 2y = −2
16. (−6, 4), (3, 1)
Solution: To write the equation, we must find the slope of this line.
m=
y2 − y1
1−4
−3
1
=
=
=− .
x2 − x1
3 − (−6)
9
3
We may choose either point to write the equation.
1
y − 4 = − (x − (−6))
3
1
y − 4 = − (x + 6)
3
1
y−4= − x−2
3
(a)
1
y = − x+2
3
To write the equation in standard form, we must move the x’s to the left side, eliminate any fractions, and
make the coefficient of the x term positive.
1
y = − x+2
3
1
x+y = 2
3
(b)
x + 3y = 6
Linear Equations Solutions
Intermediate Algebra, Joseph Lee
17. (−4, 0), (2, −12)
Solution: To write the equation, we must find the slope of this line.
y2 − y1
−12 − 0
−12
=
=
= −2.
x2 − x1
2 − (−4)
6
m=
We may choose either point to write the equation.
y − 0 = −2(x − (−4))
y = −2(x + 4)
y = −2x − 8
(a)
To write the equation in standard form, we must move the x’s to the left side, eliminate any fractions, and
make the coefficient of the x term positive.
y = −2x − 8
2x + y = −8
(b)
18. (5, 2), (−3, −2)
Solution: To write the equation, we must find the slope of this line.
m=
−4
1
−2 − 2
y2 − y1
=
= .
=
x2 − x1
−3 − 5
−8
2
We may choose either point to write the equation.
(a)
y−2=
1
(x − 5)
2
y−2=
1
5
x−
2
2
y=
1
1
x−
2
2
To write the equation in standard form, we must move the x’s to the left side, eliminate any fractions, and
make the coefficient of the x term positive.
y=
1
1
x−
2
2
1
1
− x+y = −
2
2
(b)
x − 2y = 1
19. (−7, 6), (2, 3)
Solution: To write the equation, we must find the slope of this line.
m=
y2 − y1
3−6
−3
1
=
=
=− .
x2 − x1
2 − (−7)
9
3
Linear Equations Solutions
Intermediate Algebra, Joseph Lee
We may choose either point to write the equation.
1
y − 6 = − (x − (−7))
3
1
y − 6 = − (x + 7)
3
1
7
y−6= − x−
3
3
(a)
1
11
y = − x+
3
3
To write the equation in standard form, we must move the x’s to the left side, eliminate any fractions, and
make the coefficient of the x term positive.
11
1
y = − x+
3
3
1
11
x+y =
3
3
(b)
x + 3y = 11
(20-25) Determine whether the lines are parallel, perpendicular, or neither.
1
x−2
3
1
y = x+4
3
20. y =
Solution: The first line has a slope m1 = 13 , and the second line has a slope m2 = 13 . Since m1 = m2 , the
lines are parallel.
21. y = −2x + 4
1
y =− x−5
2
Solution: The first line has a slope m1 = −2, and the second line has a slope m2 = − 12 . Since m1 6= m2
and m1 · m2 6= −1, the lines are neither parallel nor perpendicular.
22. 2x − 6y = 5
3x + y = 5
Solution: First, we will write both equations in slope-intercept form.
(L1 )
2x − 6y = 5
−6y = −2x + 5
1
5
x−
3
6
3x + y = 5
y=
(L2 )
y = −3x + 5
Linear Equations Solutions
Intermediate Algebra, Joseph Lee
The first line has a slope m1 = 13 , and the second line has a slope m2 = −3. Since m1 · m2 = −1, the lines
are perpendicular.
23. 3x − 5y = 15
5x − 3y = 10
Solution: First, we will write both equations in slope-intercept form.
(L1 )
3x − 5y = 15
−5y = −3x + 15
y=
(L2 )
3
x−3
5
5x − 3y = 10
−3y = −5x + 10
y=
10
5
x−
3
3
The first line has a slope m1 = 35 , and the second line has a slope m2 = 35 . Since m1 6= m2 and m1 · m2 6= −1,
the lines are neither parallel nor perpendicular.
24. 4x − 2y = 8
−6x + 3y = 4
Solution: First, we will write both equations in slope-intercept form.
(L1 )
4x − 2y = 8
−2y = −4x + 8
y = 2x − 4
(L2 )
− 6x + 3y = 4
3y = 6x + 4
4
3
The first line has a slope m1 = 2, and the second line has a slope m2 = 2. Since m1 = m2 , the lines are
parallel.
y = 2x +
25. x = 3
y = −1
Solution: The equation x = 3 is a vertical line, and the equation y = −1 is a horizontal line. Any two
vertical and horizontal lines are perpendicular.
(26-30) Write the equation of the line that is parallel to the given line and passes through the given point. Write the
equation in the following:
(a) slope-intercept form
(b) standard form.
Linear Equations Solutions
Intermediate Algebra, Joseph Lee
26. (3, 2), y = 3x − 4
Solution: Since the line is parallel to y = 3x − 4, we know the slope must be the same: m = 3.
y − y1 = m(x − x1 )
y − 2 = 3(x − 3)
y − 2 = 3x − 9
(a)
y = 3x − 7
−3x + y = −7
(b)
3x − y = 7
2
27. (−6, 2), y = − x + 7
3
2
Solution: Since the line is parallel to y = − 32 x + 7, we know the slope must be the same: m = − .
3
y − y1 = m(x − x1 )
2
y − 2 = − (x − (−6))
3
2
y − 2 = − (x + 6)
3
2
y−2= − x−4
3
(a)
2
y = − x−2
3
2
x + y = −2
3
(b)
2x + 3y = −6
28. (−3, −5), 4x − y = 8
Solution: First, we will write 4x − y = 8 in slope-intercept form.
4x − y = 8
−y = −4x + 8
y = 4x − 8
Linear Equations Solutions
Intermediate Algebra, Joseph Lee
Since the line is parallel to y = 4x − 8, we know the slope must be the same: m = 4.
y − y1 = m(x − x1 )
y − (−5) = 4(x − (−3))
y + 5 = 4(x + 3)
y + 5 = 4x + 12
(a)
y = 4x + 7
−4x + y = 7
(b)
4x − y = −7
29. (4, −5), 6x + 4y = 4
Solution: First, we will write 4x − y = 8 in slope-intercept form.
6x + 4y = 4
4y = −6x + 4
3
y = − x+1
2
Since the line is parallel to y = − 32 x + 1, we know the slope must be the same: m = − 23 .
y − y1 = m(x − x1 )
3
y − (−5) = − (x − 4)
2
3
y+5= − x+6
2
(a)
3
y = − x+1
2
3
x+y = 1
2
(b)
3x + 2y = 2
30. (3, −4), 3x − 5y = 15
Solution: First, we will write 3x − 5y = 15 in slope-intercept form.
3x − 5y = 15
−5y = −3x + 15
3
y = x−3
5
Linear Equations Solutions
Intermediate Algebra, Joseph Lee
Since the line is parallel to y = 35 x − 3, we know the slope must be the same: m = 35 .
y − y1 = m(x − x1 )
y − (−4) =
y+4=
(a)
y=
3
(x − 3)
5
3
9
x−
5
5
3
29
x−
5
5
3
29
− x+y = −
5
5
(b)
3x − 5y = 29
(31-35) Write the equation of the line that is perpendicular to the given line and passes through the given point.
Write the equation in the following:
(a) slope-intercept form
(b) standard form.
31. (3, 2), y = 3x − 4
Solution: Since the line is perpendicular to y = 3x − 4, we know the slope must be its opposite reciprocal:
m = − 31 .
y − y1 = m(x − x1 )
1
y − 2 = − (x − 3)
3
1
y−2= − x+1
3
(a)
1
y = − x+3
3
1
x+y = 3
3
(b)
x + 3y = 9
2
32. (−6, 2), y = − x + 7
3
Solution: Since the line is perpendicular to y = − 32 x + 7, we know the slope must be its opposite reciprocal:
Linear Equations Solutions
Intermediate Algebra, Joseph Lee
m = 32 .
y − y1 = m(x − x1 )
(a)
y−2=
3
(x − (−6))
2
y−2=
3
(x + 6)
2
y−2=
3
x+9
2
y=
3
x + 11
2
3
− x + y = 11
2
(b)
3x − 2y = −22
33. (3, −6), 5x + 2y = 10
Solution: First, we will write 5x + 2y = 10 in slope-intercept form.
5x + 2y = 10
2y = −5x + 10
5
y = − x+5
2
Since the line is perpendicular to y = − 25 x + 5, we know the slope must be its opposite reciprocal: m = 52 .
y − y1 = m(x − x1 )
y − (−6) =
y+6=
(a)
y=
2
(x − 3)
5
2
6
x−
5
5
2
36
x−
5
5
36
2
− x+y = −
5
5
(b)
2x − 5y = 36
Linear Equations Solutions
Intermediate Algebra, Joseph Lee
34. (−4, 6), 2x − 8y = 16
Solution: First, we will write 2x − 8y = 16 in slope-intercept form.
2x − 8y = 16
−8y = −2x + 16
y=
1
x−2
4
Since the line is perpendicular to y = 41 x − 2, we know the slope must be its opposite reciprocal: m = −4.
y − y1 = m(x − x1 )
y − 6 = −4(x − (−4))
y − 6 = −4(x + 4)
y − 6 = −4x − 16
(a)
(b)
y = −4x − 10
4x + y = −10
35. (5, 1), 10x − 5y = 15
Solution: First, we will write 10x − 5y = 15 in slope-intercept form.
10x − 5y = 15
−5y = −10x + 15
y = 2x − 3
Since the line is perpendicular to y = 2x − 3, we know the slope must be its opposite reciprocal: m = − 12 .
y − y1 = m(x − x1 )
1
y − 1 = − (x − 5)
2
1
5
y−1= − x+
2
2
(a)
7
1
y = − x+
2
2
1
7
x+y =
2
2
(b)
x + 2y = 7
(36-37) Write the equation of the line with the given properties.
36. parallel to y = −3 through the point (5, 2)
Linear Equations Solutions
Intermediate Algebra, Joseph Lee
Solution: Since y = −3 is a horizontal line, any line parallel to it is also horizontal. So the horizontal line
that passes through (5, 2) is the line give by
y = 2.
37. perpendicular to y = 1 through the point (−3, −5)
Solution: Since y = 1 is a horizontal line, any line perpendicular to it is vertical. So the vertical line that
passes through (−3, 5) is the line give by
x = −3.