# Honors Algebra 2: Extra Practice on Linear Equations Worksheet 1

• Class: Honors Algebra 2
• Author: Peter Atlas
• Text: Algebra and Trigonometry: Structure and Method, Brown

1. Given $$y = 4x + 2$$
1. Find the line's x-intercept(s)
2. Solution $$\displaystyle \left( -\frac{1}{2}, 0 \right)$$
3. Find the line's y-intercept(s)
4. Solution (0, 2)
5. Find the line's slope.
6. Solution 4
7. Write the equation of the line in standard form.
8. Solution $$4x - y = -2$$
2. Given $$\displaystyle y - 6 = \frac{3}{4}(x + 1)$$
1. Find the line's x-intercept(s)
2. Solution (-9, 0)
3. Find the line's y-intercept(s)
4. Solution $$\displaystyle (0, \frac{27}{4})$$
5. Find the line's slope.
6. Solution $$\displaystyle \frac{3}{4}$$
7. Write the equation of the line in standard form.
8. Solution $$3x - 4y -27$$
3. Given $$\displaystyle -\frac{x}{8} + \frac{2y}{3} = 1$$
1. Find the line's x-intercept(s)
2. Solution (-8, 0)
3. Find the line's y-intercept(s)
4. Solution $$\displaystyle \left( 0, \frac{3}{2}\right)$$
5. Find the line's slope.
6. Solution $$\displaystyle \frac{3}{16}$$
7. Write the equation of the line in standard form.
8. Solution $$3x - 16y = -24$$
4. Given $$3(2x - 7) = 6x + 9y$$
1. Find the line's x-intercept(s)
2. Solution There is no x-intercept
3. Find the line's y-intercept(s)
4. Solution $$\displaystyle \left( 0, -\frac{7}{3} \right)$$
5. Find the line's slope.
6. Solution 0
7. Write the equation of the line in standard form.
8. Solution $$0x + 3y = -7$$
5. Given $$-9x + 6y = -2$$
1. Find the line's x-intercept(s)
2. Solution $$\displaystyle \left( \frac{9}{2}, 0 \right)$$
3. Find the line's y-intercept(s)
4. Solution $$\displaystyle \left( 0, -\frac{1}{3} \right)$$
5. Find the line's slope.
6. Solution $$\displaystyle \frac{3}{2}$$
7. Write the equation of the line in standard form.
8. Solution $$9x - 6y = 2$$
6. Find the equation of the line through the points (8, -1) and (3, -2). Answer in standard form.
7. Solution $$\displaystyle y + 1 = \frac{1}{5} \left( x - 8 \right)$$; In standard form, that's $$x - 5y = 13$$.
8. Find the equation of the line parallel to the line 5x + 8y = 7 that goes through the point (4, 0). Answer in standard form.
9. Solution $$\displaystyle y - 0= -\frac{5}{8}\left(x - 4 \right)$$; In standard form, that's $$5x + 8y = 20$$.

10. Find the equation of the perpendicular bisector of the segment through the points (5, 1) and (-2, 13). Answer in standard form.
11. Solution $$\displaystyle y - 7 = \frac{7}{12} \left( x - \frac{3}{2} \right)$$; In standard form, that's $$14x - 24y = -147$$.
12. Find the equation of the line through (12, 21) perpendicular to the line y = 3. Answer in standard form.
13. Solution $$x = 12$$. In standard form, that's $$x + 0y = 12$$
14. Find the equation of the line parallel to the y-axis that goes through the point (4, -1). Answer in standard form.
15. Solution $$x = 4$$. In standard form, that's $$x + 0y = 4$$
16. Find the equation of the line with y-intercept (0, 9) with no x-intercept. Answer in standard form.
17. Solution $$y = 9$$. In standard form, that's $$0x + y = 9$$
18. Find the equation of the line with slope 3 and x-intercept 4. Leave your answer in the easiest form.
19. Solution $$y - 0 = 3(x - 4)$$
20. Find the equation of the line with y-intercept $$\displaystyle -\frac{3}{2}$$ and the same slope as $$\displaystyle \frac{x}{2} = -2y + 1$$. Leave your answer in the easiest form.
21. Solution $$\displaystyle y = -\frac{1}{4}x - \frac{3}{2}$$
22. Find the equation of the line with slope 0 through (-2, 6). Leave your answer in the easiest form.
23. Solution $$y = 6$$
24. Find the equation of the line through (4, 9) with the same slope as x = 10. Leave your answer in the easiest form.
25. Solution $$x = 4$$
26. Find the equation of the line through (3, -1) and parallel to 4x - 3y = 16. Leave your answer in the easiest form.
27. Solution $$\displaystyle y + 1 = \frac{4}{3}(x - 3)$$
28. Find the value of $$a$$ for which the line through points P$$\displaystyle \left( 4, \frac{1}{a + 1} \right)$$ and Q(10, 3) have the same slope as the line $$-4x + y = 18$$.
29. Solution $$\displaystyle 4 = \frac{3 - \frac{1}{a + 1}}{10 - 4} = \frac{3 - \frac{1}{a + 1}}{6}$$. Multiply both sides of the equation by 6 to get: $$\displaystyle 24 = 3 - \frac{1}{a + 1}$$ (and $$a \neq -1$$). Multiply both sides by $$a + 1$$ to get $$24(a + 1) = 3(a + 1) - 1$$, so $$21(a + 1) = -1$$, or $$21a + 21 = -1$$, and $$\displaystyle a = -\frac{22}{21}$$.