Slope and y-intercept

In a linear equation, the coefficient of x is the slope. Here the slope is b-9, so b-9=1 gives b=10. The y-intercept is the constant term 3b-2. Thus, 3(10)-2=28, so 28 is correct. Getting 30 would mean forgetting to subtract 2. Exam tip: put x=0 to find the y-intercept.

Use y = mx + c to identify the slope and y-intercept. Rearrange 3x - 4y + 12 = 0 by moving the other terms: -4y = -3x - 12. Dividing by -4 gives y = (3/4)x + 3. Thus m = 3/4 and c = 3, so option A is correct. The negative coefficient of y must be handled carefully; dividing the entire equation by -4 changes both signs. Option B keeps the wrong signs, option C confuses the coefficient of y with the slope, and option D uses coefficients from the unrearranged equation rather than the slope-intercept form.

In the form \(y=mx+c\), \(m\) is the slope and \(c\) is the \(y\)-intercept. The slopes here are \(t+2\) and \(9\). Equal slopes give \(t+2=9\), so \(t=7\). The \(y\)-intercept of the second line is \(2t-1=2(7)-1=13\). Although \(11\) is the value of \(t+2\), it is not the \(y\)-intercept. Exam tip: find the parameter from the slopes first, then substitute it into the constant term.

In the form \(y=mx+c\), \(c\) is the \(y\)-intercept. The first line has \(y\)-intercept \(v-5\), while the second has \(y\)-intercept \(6\). Since they are equal, \(v-5=6\), so \(v=11\). The slope of the second line is the coefficient of \(x\), namely \(v+2\); hence it is \(11+2=13\). Option 11 is the value of \(v\), not the slope. Exam tip: in \(y=mx+c\), the coefficient of \(x\) is always the slope.

The slope is \(-\frac{4}{3}\). Thus, when \(x\) increases by 3 units, \(y\) decreases by 4 units. For a decrease of 28 in \(y\), \(\frac{4}{3}\Delta x=28\). Hence, \(\Delta x=28\times\frac{3}{4}=21\). Therefore, the correct answer is 21. An increase of 18 would decrease \(y\) by only 24. Exam tip: a negative slope means that \(y\) decreases as \(x\) increases.

In \(y=mx+c\), a falling line needs \(m<0\), while an intercept above the origin needs \(c>0\). Option A has \(m=-2\) and \(c=5\). Option C falls too, but its y-intercept is below zero. Exam tip: check the signs of \(m\) and \(c\) separately.

In the form (y=mx+c), m represents the slope. Given (m-c=16) and (c=-9), we get (m-(-9)=16), or (m+9=16). Hence, (m=7), so 7 is the correct option. If m were 8, then (m-c=8-(-9)=17), not 16. Exam tip: subtracting a negative value changes it to addition.

In the form (y=mx+c), the y-intercept is c. Substituting m=-3 into 3c+m=18 gives 3c-3=18. Hence, 3c=21 and c=7. Therefore, the y-intercept is 7. Option 6 may result from incorrectly ignoring the -3 term. Exam tip: to find the y-intercept, determine the value of c.

The required form is y = mx + c, so first isolate y. From 12x + 4y = 32, subtract 12x from both sides to get 4y = -12x + 32. Dividing by 4 gives y = -3x + 8. Comparing this with y = mx + c, the coefficient of x is m = -3 and the constant is c = 8. Therefore option A is correct. Option B loses the negative sign produced when 12x moves across the equality sign. Options C and D quote original coefficients or signs instead of the values in the rearranged equation, so they do not represent the slope-intercept form.

In the form y = mx + c, the coefficient of x, m, is the slope. Here, the slope is 2k - 5. So, 2k - 5 = 13, which gives 2k = 18 and k = 9. If k were 8, the slope would be 11, not 13. Exam tip: To identify slope, look for the coefficient of x.

First divide the equation by 5: \(y=3x-8\). In the form \(y=mx+c\), the constant term \(c\) is the \(y\)-intercept. Therefore, the \(y\)-intercept is \(-8\), and the line meets the \(y\)-axis at \((0,-8)\). Although \(-40\) is the constant term in the original equation, it is not the intercept until the equation is written in slope-intercept form. Exam tip: set \(x=0\) to find the \(y\)-intercept directly.

Since the line \(y=-5x+c\) passes through \((4,-11)\), substitute \(x=4\) and \(y=-11\): \(-11=-5(4)+c=-20+c\). Hence, \(c=9\). Therefore, the y-coordinate of the \(y\)-intercept is \(9\), so the correct option is \(9\); the line meets the y-axis at \((0,9)\). Exam tip: in \(y=mx+c\), \(c\) directly gives the y-intercept value.

Substitute the point \((-3,6)\) into the line: \(6=m(-3)-9\). Hence, \(6=-3m-9\), so \(15=-3m\) and \(m=-5\). Therefore, the slope is \(-5\). The value \(5\) may result from a sign error; the negative \(x\)-coordinate gives \(-3m\). Exam tip: substitute the given point carefully, keeping the signs intact, and then solve for the slope.

In a line of the form \(x=a\), the x-coordinate remains fixed, so the line is vertical and parallel to the y-axis. \(y=4\) is horizontal instead. Exam tip: a constant \(x\) indicates a vertical line.

In \(y=mx+c\), \(m\) is the slope and \(c\) is the y-intercept. Option A has \(m=-2\), which is negative, and \(c=4\). Option B has a positive slope. Exam tip: check the coefficient of \(x\) first.

The line \(x=5\) is parallel to the y-axis. For a y-intercept, put \(x=0\), but \(0=5\) is impossible, so it never meets the y-axis. In contrast, \(y=5\) has y-intercept 5. Exam tip: \(x=a\), for \(a\ne0\), has no y-intercept.

In the form \(y=mx+c\), the coefficient of \(x\) is the slope \(m\). Here, the slope is \(3p+1\). So, \(3p+1=22\), giving \(3p=21\) and \(p=7\). If \(p=6\), the slope would be \(3(6)+1=19\), not 22. Exam tip: identify the coefficient of \(x\) before equating slopes.

The line is in the form y = mx + c, where the slope m = r + 6 and the y-intercept c = r - 3. A line passing through the origin (0, 0) must have y-intercept 0. Hence, r - 3 = 0, so r = 3. Therefore, the slope is r + 6 = 3 + 6 = 9. Option 3 is the value of r, not the slope. Exam tip: For y = mx + c to pass through the origin, set c = 0 directly.

In y = mx + c, m is the slope and c is the y-intercept. Here m = -2, so y decreases by 2 when x increases by 1; c = 6. Exam tip: always check the sign of the slope.

To find the y-intercept, put x=0 because every point on the y-axis has x-coordinate 0. Then -3y+15=0, so -3y=-15 and y=5. Hence, the y-intercept is 5. Taking -5 does not satisfy the equation. Exam tip: for a y-intercept, always set x=0.