Question 185456
# 1


Let's solve this system by elimination




Start with the given system of equations:

{{{system(3x-2y=26,-7x+3y=-49)}}}



{{{3(3x-2y)=3(26)}}} Multiply the both sides of the first equation by 3.



{{{9x-6y=78}}} Distribute and multiply.



{{{2(-7x+3y)=2(-49)}}} Multiply the both sides of the second equation by 2.



{{{-14x+6y=-98}}} Distribute and multiply.



So we have the new system of equations:

{{{system(9x-6y=78,-14x+6y=-98)}}}



Now add the equations together. You can do this by simply adding the two left sides and the two right sides separately like this:



{{{(9x-6y)+(-14x+6y)=(78)+(-98)}}}



{{{(9x+-14x)+(-6y+6y)=78+-98}}} Group like terms.



{{{-5x+0y=-20}}} Combine like terms.



{{{-5x=-20}}} Simplify.



{{{x=(-20)/(-5)}}} Divide both sides by {{{-5}}} to isolate {{{x}}}.



{{{x=4}}} Reduce.



------------------------------------------------------------------



{{{9x-6y=78}}} Now go back to the first equation.



{{{9(4)-6y=78}}} Plug in {{{x=4}}}.



{{{36-6y=78}}} Multiply.



{{{-6y=78-36}}} Subtract {{{36}}} from both sides.



{{{-6y=42}}} Combine like terms on the right side.



{{{y=(42)/(-6)}}} Divide both sides by {{{-6}}} to isolate {{{y}}}.



{{{y=-7}}} Reduce.



So the solutions are {{{x=4}}} and {{{y=-7}}}.



Which form the ordered pair *[Tex \LARGE \left(4,-7\right)].



This means that the system is consistent and independent.



Notice when we graph the equations, we see that they intersect at *[Tex \LARGE \left(4,-7\right)]. So this visually verifies our answer.



{{{drawing(500,500,-6,14,-17,3,
grid(1),
graph(500,500,-6,14,-17,3,(26-3x)/(-2),(-49+7x)/(3)),
circle(4,-7,0.05),
circle(4,-7,0.08),
circle(4,-7,0.10)
)}}} Graph of {{{3x-2y=26}}} (red) and {{{-7x+3y=-49}}} (green) 



================================================================


# 2



Let's solve this system by elimination



Start with the given system of equations:

{{{system(4x-5y=14,-12x+15y=-42)}}}



{{{3(4x-5y)=3(14)}}} Multiply the both sides of the first equation by 3.



{{{12x-15y=42}}} Distribute and multiply.



So we have the new system of equations:

{{{system(12x-15y=42,-12x+15y=-42)}}}



Now add the equations together. You can do this by simply adding the two left sides and the two right sides separately like this:



{{{(12x-15y)+(-12x+15y)=(42)+(-42)}}}



{{{(12x+-12x)+(-15y+15y)=42+-42}}} Group like terms.



{{{0x+0y=0}}} Combine like terms.



{{{0=0}}}Simplify.



Since {{{0=0}}} is <font size="4"><b>always</b></font> true, this means that there are an infinite number of solutions. 



So the system is consistent and dependent.



================================================================


# 3





Start with the given system of equations:

{{{system(-2x+6y=19,10x-30y=-15)}}}



{{{5(-2x+6y)=5(19)}}} Multiply the both sides of the first equation by 5.



{{{-10x+30y=95}}} Distribute and multiply.



So we have the new system of equations:

{{{system(-10x+30y=95,10x-30y=-15)}}}



Now add the equations together. You can do this by simply adding the two left sides and the two right sides separately like this:



{{{(-10x+30y)+(10x-30y)=(95)+(-15)}}}



{{{(-10x+10x)+(30y+-30y)=95+-15}}} Group like terms.



{{{0x+0y=80}}} Combine like terms.



{{{0=80}}}Simplify.



Since {{{0=80}}} is <font size="4"><b>never</b></font> true, this means that there are no solutions. 



So the system is inconsistent.