Question 192881
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Let us make <font color=blue>x</font>= the number of hours Bill works
And, <font color=blue>y</font>= dollars he received for <font color=blue>x</font> hours he worked.


Now, if x =8, then  y =50 right?


We know line Eqn is in the format, Ax + By = C (Std. Form)
Let {{{system(A=1,B=1)}}}


Then, {{{(1)x+(1)y=c}}}
But we know, {{{x=8}}} when {{{y=50}}}.  So our Line Eqn will be:


{{{x=y}}} plus {{{constant=c}}}


Either {{{x=y+c}}}{{{(red(EQN1))}}} or vice versa {{{y=x+c}}}{{{(red(EQN2))}}}.


Substituting {{{x=8}}} and {{{y=50}}} we can get {{{c}}}: (Use {{{red(EQN2)}}}), either eqn will be the same anyway when you substitute the values.


{{{highlight(50)=highlight(8)+c}}}
{{{50-8=c}}}
{{{red(42=c)}}}


It follows ---> y = x + c  ----> <font color=blue>y = x + 42</font>,or <font color=blue>-x + y =42</font>, Std. Form ( Answer )


As the problem says, if x = 8 hours, then y = $50.
Just by sustituting:
{{{-highlight(8)+highlight(50)=42}}}
{{{42=42}}}, good.



Or seeing the graph,

{{{drawing(500,500,-6,25,-10,100,graph(500,500,-6,25,-10,100,x+42),green(arrow(17,60,16,68)),green(locate(14,72,-x+y=42)),blue(circle(8,50,.25)),green(line(8,0,8,50)),green(line(0,50,8,50)),red(locate(10,-5,HOURS)),red(locate(-5,90,DOLLARS)))}}} ----> You can see, if Bill works 8 hours, he gets $50. The intersection is on our Line Eqn <font color=blue>-x + y = 42</font>. Try 20 hours?


Thank you,
Jojo</font>