Question 174259
# 1



{{{-2x+3y=3}}} Start with the given equation.



{{{3y=3+2x}}} Add {{{2x}}} to both sides.



{{{3y=2x+3}}} Rearrange the terms.



{{{y=(2x+3)/(3)}}} Divide both sides by {{{3}}} to isolate y.



{{{y=((2)/(3))x+(3)/(3)}}} Break up the fraction.



{{{y=(2/3)x+1}}} Reduce.



So the equation {{{y=(2/3)x+1}}} is now in slope intercept form {{{y=mx+b}}} where the slope is {{{m=2/3}}} and the y-intercept is {{{b=1}}} note: the y-intercept is the point *[Tex \LARGE \left(0,1\right)]




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# 2





First let's find the slope of the line through the points *[Tex \LARGE \left(-3,-4\right)] and *[Tex \LARGE \left(0,5\right)]



{{{m=(y[2]-y[1])/(x[2]-x[1])}}} Start with the slope formula.



{{{m=(5--4)/(0--3)}}} Plug in {{{y[2]=5}}}, {{{y[1]=-4}}}, {{{x[2]=0}}}, and {{{x[1]=-3}}}



{{{m=(9)/(0--3)}}} Subtract {{{-4}}} from {{{5}}} to get {{{9}}}



{{{m=(9)/(3)}}} Subtract {{{-3}}} from {{{0}}} to get {{{3}}}



{{{m=3}}} Reduce



So the slope of the line that goes through the points *[Tex \LARGE \left(-3,-4\right)] and *[Tex \LARGE \left(0,5\right)] is {{{m=3}}}



Now let's use the point slope formula:



{{{y-y[1]=m(x-x[1])}}} Start with the point slope formula



{{{y--4=3(x--3)}}} Plug in {{{m=3}}}, {{{x[1]=-3}}}, and {{{y[1]=-4}}}



{{{y--4=3(x+3)}}} Rewrite {{{x--3}}} as {{{x+3}}}



{{{y+4=3(x+3)}}} Rewrite {{{y--4}}} as {{{y+4}}}



{{{y+4=3x+3(3)}}} Distribute



{{{y+4=3x+9}}} Multiply



{{{y=3x+9-4}}} Subtract 4 from both sides. 



{{{y=3x+5}}} Combine like terms. 



{{{y=3x+5}}} Simplify



So the equation that goes through the points *[Tex \LARGE \left(-3,-4\right)] and *[Tex \LARGE \left(0,5\right)] is {{{y=3x+5}}}



 Notice how the graph of {{{y=3x+5}}} goes through the points *[Tex \LARGE \left(-3,-4\right)] and *[Tex \LARGE \left(0,5\right)]. So this visually verifies our answer.

 {{{drawing( 500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -10, 10,3x+5),
 circle(-3,-4,0.08),
 circle(-3,-4,0.10),
 circle(-3,-4,0.12),
 circle(0,5,0.08),
 circle(0,5,0.10),
 circle(0,5,0.12)
 )}}} Graph of {{{y=3x+5}}} through the points *[Tex \LARGE \left(-3,-4\right)] and *[Tex \LARGE \left(0,5\right)]




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# 3





If you want to find the equation of line with a given a slope of {{{-5}}} which goes through the point ({{{-4}}},{{{19}}}), 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-19=(-5)(x--4)}}} Plug in {{{m=-5}}}, {{{x[1]=-4}}}, and {{{y[1]=19}}} (these values are given)



{{{y-19=(-5)(x+4)}}} Rewrite {{{x--4}}} as {{{x+4}}}



{{{y-19=-5x+(-5)(4)}}} Distribute {{{-5}}}


{{{y-19=-5x-20}}} Multiply {{{-5}}} and {{{4}}} to get {{{-20}}}


{{{y=-5x-20+19}}} Add 19 to  both sides to isolate y


{{{y=-5x-1}}} Combine like terms {{{-20}}} and {{{19}}} to get {{{-1}}} 

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Answer:



So the equation of the line with a slope of {{{-5}}} which goes through the point ({{{-4}}},{{{19}}}) is:


{{{y=-5x-1}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=-5}}} and the y-intercept is {{{b=-1}}}


Notice if we graph the equation {{{y=-5x-1}}} and plot the point ({{{-4}}},{{{19}}}),  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, -13, 5, -2, 21,
graph(500, 500, -13, 5, -2, 21,(-5)x+-1),
circle(-4,19,0.12),
circle(-4,19,0.12+0.03)
) }}} Graph of {{{y=-5x-1}}} through the point ({{{-4}}},{{{19}}})

and we can see that the point lies on the line. Since we know the equation has a slope of {{{-5}}} and goes through the point ({{{-4}}},{{{19}}}), this verifies our answer.