Question 126330
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<a href="#One">Finding the equation and graphing</a>
<a href="#Two">Part A</a>
<a href="#Three">Part B</a>
<a href="#Four">Part C</a>



<a name="One">


Let x=mass of loonies (g) and y= # of loonies



When {{{x=21}}}, {{{y=3}}}. So we have the first point (21,3)



When {{{x=63}}}, {{{y=9}}}. So we have the second point (63,9)




So let's find the equation of the line through the points (21,3) and (63,9)



First lets find the slope through the points ({{{21}}},{{{3}}}) and ({{{63}}},{{{9}}})


{{{m=(y[2]-y[1])/(x[2]-x[1])}}} Start with the slope formula (note: *[Tex \Large \left(x_{1},y_{1}\right)] is the first point ({{{21}}},{{{3}}}) and  *[Tex \Large \left(x_{2},y_{2}\right)] is the second point ({{{63}}},{{{9}}}))


{{{m=(9-3)/(63-21)}}} Plug in {{{y[2]=9}}},{{{y[1]=3}}},{{{x[2]=63}}},{{{x[1]=21}}}  (these are the coordinates of given points)


{{{m= 6/42}}} Subtract the terms in the numerator {{{9-3}}} to get {{{6}}}.  Subtract the terms in the denominator {{{63-21}}} to get {{{42}}}

  


{{{m=1/7}}} Reduce

  

So the slope is

{{{m=1/7}}}


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Now let's use the point-slope formula to find the equation of the line:




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


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


{{{y-3=(1/7)(x-21)}}} Plug in {{{m=1/7}}}, {{{x[1]=21}}}, and {{{y[1]=3}}} (these values are given)



{{{y-3=(1/7)x+(1/7)(-21)}}} Distribute {{{1/7}}}


{{{y-3=(1/7)x-3}}} Multiply {{{1/7}}} and {{{-21}}} to get {{{-21/7}}}. Now reduce {{{-21/7}}} to get {{{-3}}}


{{{y=(1/7)x-3+3}}} Add {{{3}}} to  both sides to isolate y


{{{y=(1/7)x+0}}} Combine like terms {{{-3}}} and {{{3}}} to get {{{0}}} 


{{{y=(1/7)x}}} Remove the zero term





So the equation of the line which goes through the points ({{{21}}},{{{3}}}) and ({{{63}}},{{{9}}})  is: {{{y=(1/7)x}}}


The equation is now in {{{y=mx+b}}} form (which is slope-intercept form) where the slope is {{{m=1/7}}} and the y-intercept is {{{b=0}}}


Notice if we graph the equation {{{y=(1/7)x}}} and plot the points ({{{21}}},{{{3}}}) and ({{{63}}},{{{9}}}),  we get this: (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, 15, 75, -3, 15,
graph(500, 500, 15, 75, -3, 15,(1/7)x+0),
circle(21,3,0.12),
circle(21,3,0.12+0.03),
circle(63,9,0.12),
circle(63,9,0.12+0.03)
) }}} Graph of {{{y=(1/7)x}}} through the points ({{{21}}},{{{3}}}) and ({{{63}}},{{{9}}})


Notice how the two points lie on the line. This graphically verifies our answer.



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<a name="Two">

a)


"what is the mass of 33 loonies?"



To find the total mass (which is x) simply plug in {{{y=33}}}



{{{y=(1/7)x}}} Start with the given equation


{{{33=(1/7)x}}} Plug in {{{y=33}}}



{{{33*7=x}}} Multiply both sides by 7 to isolate x



{{{231=x}}} Multiply 



{{{x=231}}}



So 33 loonies weigh 231 g


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<a name="Three">


b)


"What is teh value of 133g of loonies?"



First let's find the number of loonies (which is y)



{{{y=(1/7)x}}} Start with the given equation



{{{y=(1/7)133}}} Plug in {{{x=133}}}



{{{y=133/7}}} Multiply



{{{y=19}}} Divide



So there are 19 loonies that weigh 133 g. So multiply the value of one loonie by 19 to find the total value. 



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<a name="Four">


c)


{{{y=(1/7)x}}} Start with the given equation



{{{y=(1/7)152}}} Plug {{{x=152}}}



{{{y=152/7}}} Multiply



{{{y=21.714}}} Divide



Since you cannot have 21.714 loonies, there must be other types of coins.