Question 166670

{{{6x+5y=3}}} Start with the first equation.



{{{5y=3-6x}}} Subtract {{{6x}}} from both sides.



{{{5y=-6x+3}}} Rearrange the terms.



{{{y=(-6x+3)/(5)}}} Divide both sides by {{{5}}} to isolate y.



{{{y=((-6)/(5))x+(3)/(5)}}} Break up the fraction.



{{{y=-(6/5)x+3/5}}} Reduce.



So we can see that the equation {{{y=-(6/5)x+3/5}}} has a slope {{{m=-6/5}}} and a y-intercept {{{b=3/5}}}.



{{{5x-6y=8}}} Now move onto the second equation.



{{{-6y=-5x+8}}} Rearrange the terms.



{{{y=(-5x+8)/(-6)}}} Divide both sides by {{{-6}}} to isolate y.



{{{y=((-5)/(-6))x+(8)/(-6)}}} Break up the fraction.



{{{y=(5/6)x-4/3}}} Reduce.



So we can see that the equation {{{y=(5/6)x-4/3}}} has a slope {{{m=5/6}}} and a y-intercept {{{b=-4/3}}}.



So the slope of the first line is {{{m=-6/5}}} and the slope of the second line is {{{m=5/6}}}.



Notice how the slope of the second line {{{m=5/6}}} is simply the negative reciprocal of the slope of the first line {{{m=-6/5}}}.



In other words, if you flip the fraction of the second slope and change its sign, you'll get the first slope. So this means that the graphs of {{{y=-(6/5)x+3/5}}} and {{{y=(5/6)x-4/3}}} are perpendicular lines.



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Answer:


Consequently, this means that the graphs of {{{6x+5y=3}}} and {{{5x-6y=8}}} are perpendicular lines.