Question 174264
I'll do the first two to get you going in the right direction


# 1




If you want to find the equation of line with a given a slope of {{{-3/4}}} which goes through the point ({{{0}}},{{{4}}}), you can simply use the point-slope formula to find the equation:



---Point-Slope Formula---
{{{y-y[1]=m(x-x[1])}}} where {{{m}}} is the slope, and *[Tex \Large \left(x_{1},y_{1}\right)] is the given point


So lets use the Point-Slope Formula to find the equation of the line


{{{y-4=(-3/4)(x-0)}}} Plug in {{{m=-3/4}}}, {{{x[1]=0}}}, and {{{y[1]=4}}} (these values are given)



{{{y-4=(-3/4)x+(-3/4)(-0)}}} Distribute {{{-3/4}}}


{{{y-4=(-3/4)x+0}}} Multiply {{{-3/4}}} and {{{-0}}} to get {{{0}}}


{{{y=(-3/4)x+0+4}}} Add 4 to  both sides to isolate y


{{{y=(-3/4)x+4}}} Combine like terms {{{0}}} and {{{4}}} to get {{{4}}} 

------------------------------------------------------------------------------------------------------------

Answer:



So the equation of the line with a slope of {{{-3/4}}} which goes through the point ({{{0}}},{{{4}}}) is:


{{{y=(-3/4)x+4}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=-3/4}}} and the y-intercept is {{{b=4}}}


Notice if we graph the equation {{{y=(-3/4)x+4}}} and plot the point ({{{0}}},{{{4}}}),  we get (note: if you need help with graphing, check out this <a href=http://www.algebra.com/algebra/homework/Linear-equations/graphing-linear-equations.solver>solver<a>)


{{{drawing(500, 500, -9, 9, -5, 13,
graph(500, 500, -9, 9, -5, 13,(-3/4)x+4),
circle(0,4,0.12),
circle(0,4,0.12+0.03)
) }}} Graph of {{{y=(-3/4)x+4}}} through the point ({{{0}}},{{{4}}})

and we can see that the point lies on the line. Since we know the equation has a slope of {{{-3/4}}} and goes through the point ({{{0}}},{{{4}}}), this verifies our answer.



<hr>


# 2




{{{y^2-11y=-24}}} Start with the given equation.



{{{y^2-11y+24=0}}} Add 24 to both sides.



Notice we have a quadratic equation in the form of {{{ay^2+by+c}}} where {{{a=1}}}, {{{b=-11}}}, and {{{c=24}}}



Let's use the quadratic formula to solve for y



{{{y = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{y = (-(-11) +- sqrt( (-11)^2-4(1)(24) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=-11}}}, and {{{c=24}}}



{{{y = (11 +- sqrt( (-11)^2-4(1)(24) ))/(2(1))}}} Negate {{{-11}}} to get {{{11}}}. 



{{{y = (11 +- sqrt( 121-4(1)(24) ))/(2(1))}}} Square {{{-11}}} to get {{{121}}}. 



{{{y = (11 +- sqrt( 121-96 ))/(2(1))}}} Multiply {{{4(1)(24)}}} to get {{{96}}}



{{{y = (11 +- sqrt( 25 ))/(2(1))}}} Subtract {{{96}}} from {{{121}}} to get {{{25}}}



{{{y = (11 +- sqrt( 25 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{y = (11 +- 5)/(2)}}} Take the square root of {{{25}}} to get {{{5}}}. 



{{{y = (11 + 5)/(2)}}} or {{{y = (11 - 5)/(2)}}} Break up the expression. 



{{{y = (16)/(2)}}} or {{{y =  (6)/(2)}}} Combine like terms. 



{{{y = 8}}} or {{{y = 3}}} Simplify. 



So the answers are {{{y = 8}}} or {{{y = 3}}}