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Question
ws: standard form to slope-intercept form
write the slope - intercept form of the equation of each line.
1)
- 10x = y + 5
- x + 6y = 24
- 2x - 3y = -15
- x + 4y = 4
- 7x + 2y = 24
- 3x - 2y = 4
- 3x + 7y = 35
- -2y = 2 (with options: a) $y = \frac{7}{2}x - 1$, b) $y = x - \frac{7}{2}$, c) $y = -x - \frac{7}{2}$, d) $y = -\frac{7}{2}x - 1$)
- 8x - 3y = 18 (with options: a) $y = -\frac{4}{3}x - 6$, b) $y = \frac{4}{3}x - 6$, c) $y = \frac{2}{3}x - 6$, d) $y = \frac{8}{3}x - 6$)
Let's solve these problems one by one. The slope - intercept form of a line is $y = mx + b$, where $m$ is the slope and $b$ is the y - intercept. To convert a standard form equation ($Ax+By = C$) to slope - intercept form, we need to solve the equation for $y$.
Problem 1: $8x + y=-2$
Step 1: Isolate $y$
Subtract $8x$ from both sides of the equation.
$y=-8x - 2$
Problem 2: $10x=y + 5$
Step 1: Isolate $y$
Subtract $5$ from both sides (or we can also subtract $10x$ from both sides and then multiply by - 1). Let's subtract $5$ from both sides:
$y = 10x-5$
Problem 3: $x + 6y=24$
Step 1: Isolate the term with $y$
Subtract $x$ from both sides:
$6y=-x + 24$
Step 2: Solve for $y$
Divide each term by $6$:
$y=-\frac{1}{6}x + 4$
Problem 4: $2x-3y=-15$
Step 1: Isolate the term with $y$
Subtract $2x$ from both sides:
$-3y=-2x - 15$
Step 2: Solve for $y$
Divide each term by $- 3$:
$y=\frac{2}{3}x + 5$
Problem 5: $x + 4y=4$
Step 1: Isolate the term with $y$
Subtract $x$ from both sides:
$4y=-x + 4$
Step 2: Solve for $y$
Divide each term by $4$:
$y=-\frac{1}{4}x + 1$
Problem 6: $7x + 2y=24$
Step 1: Isolate the term with $y$
Subtract $7x$ from both sides:
$2y=-7x + 24$
Step 2: Solve for $y$
Divide each term by $2$:
$y=-\frac{7}{2}x+12$
Problem 7: $3x - 2y=4$
Step 1: Isolate the term with $y$
Subtract $3x$ from both sides:
$-2y=-3x + 4$
Step 2: Solve for $y$
Divide each term by $-2$:
$y=\frac{3}{2}x-2$
Problem 8: $3x + 7y=35$
Step 1: Isolate the term with $y$
Subtract $3x$ from both sides:
$7y=-3x + 35$
Step 2: Solve for $y$
Divide each term by $7$:
$y=-\frac{3}{7}x + 5$
Problem 9: Let's assume the original equation is something like $7x-2y = 2$ (since the given part is a bit unclear, but based on the options). Let's work with $7x-2y=2$
Step 1: Isolate the term with $y$
Subtract $7x$ from both sides:
$-2y=-7x + 2$
Step 2: Solve for $y$
Divide each term by $-2$:
$y=\frac{7}{2}x - 1$
Problem 10: $8x-3y = 18$
Step 1: Isolate the term with $y$
Subtract $8x$ from both sides:
$-3y=-8x + 18$
Step 2: Solve for $y$
Divide each term by $-3$:
$y=\frac{8}{3}x-6$
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s:
- $y=-8x - 2$
- $y = 10x-5$
- $y=-\frac{1}{6}x + 4$
- $y=\frac{2}{3}x + 5$
- $y=-\frac{1}{4}x + 1$
- $y=-\frac{7}{2}x+12$
- $y=\frac{3}{2}x-2$
- $y=-\frac{3}{7}x + 5$
- If the equation is $7x - 2y=2$, then $y=\frac{7}{2}x - 1$
- $y=\frac{8}{3}x-6$ (corresponding to option D)