Slope and y-intercept

From (5y=-12x+65), (y=-\frac{12}{5}x+13). So the (y)-intercept is (13) and slope is (-\frac{12}{5}).

In the form \(y=mx+c\), the \(y\)-intercept is \(c\), because \(x=0\) on the \(y\)-axis. The first line has \(y\)-intercept \(b\), while the second has \(y\)-intercept \(15\). Since they meet the \(y\)-axis at the same point, \(b=15\). The number \(9\) is the slope of the first line, not its intercept. Exam tip: in \(y=mx+c\), identify \(c\) directly as the \(y\)-intercept.

A line in the form \(y=mx+c\) has y-intercept \(c\). In option A, \(c=3\) for both lines, while the slopes are 2 and −5. Exam tip: compare the constant terms first to identify y-intercepts.

The y-intercept is the value of y when x=0. Therefore, in y=-12x+c, we have c=9. Substituting x=-4 gives y=-12(-4)+9=48+9=57. The value 48 is only the product -12×(-4); the y-intercept 9 must also be added. Exam tip: in y=mx+c, the y-intercept is c.

The slope of the line is (k-6). Since slope = change in y / change in x, (k-6)=108/9=12. Hence, (k=18). If 17 were chosen, the slope would be 11, so an increase of 9 in x would increase y by only 99, not 108. Exam tip: in a linear equation (y=mx+c), the constant c does not affect the rate of change; only m is the slope.

The slope of the line is \(-(p+5)\). A decrease of \(72\) in \(y\) for an increase of \(6\) in \(x\) gives slope \(\frac{-72}{6}=-12\). Therefore, \(-(p+5)=-12\), so \(p+5=12\) and \(p=7\). If \(p=8\), the slope would be \(-13\), giving a decrease of \(78\) for an increase of 6 in \(x\). Exam tip: In change-based questions, first find the slope using \(\frac{\Delta y}{\Delta x}\).

The y-intercept is where the line meets the y-axis, so \(x=0\). Substituting \(x=0\) in \(y=mx+c\) gives \(y=c\), hence the point is \((0,c)\). Do not confuse it with \((c,0)\), which is related to the x-axis. Exam tip: set \(x=0\) for a y-intercept.

In the form (y=mx+c), (m) is the slope and (c) is the y-intercept. Here, (m=-8). Substituting in (m+c=6) gives (-8+c=6), so (c=14). Therefore, the y-intercept is 14. Choosing (-14) would give (-8-14=-22), which does not satisfy the given condition. Exam tip: In (y=mx+c), the constant term (c) is the y-intercept.

Since \((16,25)\) lies on the line, substitute \(x=16\) and \(y=25\): \(25=\frac{7}{8}\times16+c=14+c\). Hence, \(c=25-14=11\). Option 10 can result from an error in multiplication or subtraction. Exam tip: for \(y=mx+c\), use \(c=y-mx\) after substituting the given point.

Writing the given line as \(y=\frac{2}{3}x+\frac{7}{3}\) shows that its slope is \(\frac{2}{3}\). A parallel line must have this same slope and y-intercept \(-4\). Option A rearranges to \(y=\frac{2}{3}x-4\). Exam tip: parallel lines always have equal slopes.

In the form y=mx+c, the y-intercept is c. Here, a+5=17, so a=12. Therefore, the slope is m=7a-6=7(12)-6=78. Option 76 would result if the expression were 7a-8, so it is not correct. Exam tip: first equate the constant term to the given y-intercept, then substitute the parameter into the coefficient of x.

In the form y = mx + c, the coefficient of x is the slope. Therefore, b - 15 = 5, so b = 20. The y-intercept is the constant term 4b + 2: 4(20) + 2 = 82. Hence, 82 is correct. The value 81 would result if the constant term were 4b + 1. Exam tip: first find the parameter using the slope, then substitute it into the constant term to get the y-intercept.

The direct answer is option A: slope \(\frac{9}{10}\) and y-intercept 3. Put the equation into slope-intercept form. From \(9x-10y+30=0\), subtract the x and constant terms appropriately: \(-10y=-9x-30\). Divide by \(-10\): \(y=\frac{9}{10}x+3\). Comparing with \(y=mx+c\), the slope is \(m=\frac{9}{10}\), while the y-intercept is \(c=3\). Checking directly, at \(x=0\), the original equation gives \(-10y+30=0\), so \(y=3\). Option A matches both values. Option B has both signs reversed. Option C confuses a coefficient with the slope and places \(\frac{9}{10}\) as the intercept. Option D uses unrelated original coefficients and the wrong sign for the intercept. Remember: when dividing by a negative number, both sides change signs; do not lose that sign.

In the form \(y=mx+c\), \(m\) is the slope and \(c\) is the y-intercept. Since the slopes are equal, \(t+6=16\), so \(t=10\). For the second line, the y-intercept is \(2t-9=2(10)-9=11\). Therefore, 11 is correct. The value \(-10\) is the y-intercept of the first line, not the second. Exam tip: compare the coefficients of \(x\) for slopes and use the constant term for the y-intercept.

In the form \(y=mx+c\), the \(y\)-intercept is \(c\). The first line has \(y\)-intercept \(v-11\), while the second has \(y\)-intercept \(10\). Since they are equal, \(v-11=10\), so \(v=21\). Therefore, the slope of the second line is \(v+5=21+5=26\). Option \(25\) would result from incorrectly taking \(v=20\). Exam tip: in \(y=mx+c\), \(m\) is the slope and \(c\) is the \(y\)-intercept.

The slope is \(-\frac{9}{5}\), so for every increase of 5 units in \(x\), \(y\) decreases by 9 units. For a decrease of 54 in \(y\), \(\frac{9}{5}\Delta x=54\). Hence, \(\Delta x=54\times\frac{5}{9}=30\). If 35 were chosen, \(y\) would decrease by \(63\), not 54. Exam tip: in change-based questions, the constant 22 does not affect the change; use only the slope and \(\Delta x\).

In the form \(y=mx+c\), \(m\) is the slope and \(c\) is the y-intercept. Here \(m=-3\) and \(c=5\), so the line falls from left to right and meets the y-axis at \(5\). Exam tip: the coefficient of \(x\) gives the slope.

Given c=-18 and m-c=31, we get m-(-18)=31, or m+18=31. Hence, m=13. Therefore, the slope is 13. If m were 12, then m-c=12-(-18)=30, not 31. Exam tip: subtracting a negative number is the same as adding it.

In \(y=mx+c\), \(m\) is the slope and \(c\) is the y-intercept. Here \(m=3\) and \(c=-5\). Check by putting \(x=0\): \(y=-5\). Exam tip: set \(x=0\) to find the y-intercept.