Question 203891
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Given: {{{5x+4y=12 }}}, line equation


* To find the X & Y Intercepts of the line, we let F(x)=0 to find Y-intercept, & let F(y)=0 to find X-Intercept:

F(x)=0
{{{5(0)+4y=12}}} ---> {{{4y=12}}} ---> {{{cross(4)y/cross(4)+cross(12)3/cross(4)}}}
{{{red(y=3)}}}, Y-Intercept


F(y)=0
{{{5x+4(0)=12}}} ---> {{{cross(5)x/cross(5)=12/5}}}
{{{red(x=2.4)}}}, X-Intercept


Via Slope Intercept Form, {{{y=mx+b}}}, where{{{system(m=slope, b=Yintercept)}}}


Following this form,
{{{5x+4y=12}}} ---> {{{4y=12-5x}}} ---> {{{cross(4)y/cross(4)=(12-5x)/4}}}
{{{y=12/4-5x/4}}}
{{{red(y=-(5/4)x+3)}}}


So the slope=m is {{{red(-5/4)}}}, and Y-intercept is {{{red(3)}}}. (Answer)


When x=16;
{{{y=(-5/4)(16)+3=-80/4+3=-20+3}}}
{{{red(y=-17)}}}, (Answer)

{{{drawing(500,500,-5,25,-20,10,graph(500,500,-5,25,-20,10,(-5/4)x+3),blue(circle(0,3,.15)),green(circle(16,-17,.18)),green(line(0,-17,16,-17)),green(line(16,0,16,-17)),green(arrow(9,-11,10,-10)),red(locate(6,-11,5x+4y=12)),blue(circle(2.4,0,.15)),red(locate(2.4,1,Xintercept)),red(locate(.8,3.5,Yintercept)))}}}


For the {{{Slope=m=-5/4=Y_axis/X_axis}}}.

From the Y-intercept = (0,3), we moved down 5 steps (-5), and to the right 4 steps (4).
{{{drawing(500,500,-5,25,-20,10,grid(1),graph(500,500,-5,25,-20,10,(-5/4)x+3),blue(circle(0,3,.15)),green(circle(16,-17,.18)),green(line(0,-17,16,-17)),green(line(16,0,16,-17)),green(arrow(9,-11,10,-10)),red(locate(6,-11,5x+4y=12)),blue(circle(2.4,0,.15)),green(circle(4,-2,.15)),green(circle(8,-7,.15)),green(circle(12,-12,.15)))}}} ---> See, if you keep doing that, that's where the point on the X-axis & Y-axis meet on the line.


Thank you,
Jojo</font>