Question 173267
The goal of solving ANY system of equations with 2 variables is to eliminate one variable so you can solve for the other variable. To eliminate one variable, simply "substitute" an expression in terms of the other variable. 



# 1


Note: I'm assuming that the second equation is {{{y = 2x-1}}}



{{{y = -3x+19}}} Start with the first equation



{{{2x-1 = -3x+19}}} Plug in {{{y = 2x-1}}}. In other words, replace "y" with 2x-1. Notice how the "y" term is gone.



Now we can solve for "x".



{{{2x=-3x+19+1}}} Add {{{1}}} to both sides.



{{{2x+3x=19+1}}} Add {{{3x}}} to both sides.



{{{5x=19+1}}} Combine like terms on the left side.



{{{5x=20}}} Combine like terms on the right side.



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



{{{x=4}}} Reduce. So this is the first answer.



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{{{y = 2x-1}}} Go back to the second equation



{{{y = 2(4)-1}}} Plug in {{{x=4}}}



{{{y = 8-1}}} Multiply



{{{y = 7}}} Subtract. So this is the second answer.



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Answer:



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



which form the ordered pair (4,7)



So the system is consistent and independent.



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# 2



{{{3x-2y=6 }}} Start with the second equation



{{{3x-2(x+4)=6 }}} Plug in {{{y = x +4}}}. In other words, replace "y" with x+4. Notice how the "y" term is gone.



Now we can solve for "x".



{{{3x-2x-8=6}}} Distribute.



{{{x-8=6}}} Combine like terms on the left side.



{{{x=6+8}}} Add {{{8}}} to both sides.



{{{x=14}}} Combine like terms on the right side. So this is the first answer.



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{{{y = x +4}}} Go back to the first equation



{{{y = 14 +4}}} Plug in {{{x=14}}}



{{{y = 18}}} Add. So this is the second answer.



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Answer:



So the solutions are {{{x=14}}} and {{{y = 18}}}



which form the ordered pair (14,18)



So the system is consistent and independent.



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# 3






Start with the given system of equations:


{{{system(2x-y=4,2x-y=3)}}}




Now in order to solve this system by using substitution, we need to solve (or isolate) one variable. I'm going to solve for y.





So let's isolate y in the first equation



{{{2x-y=4}}} Start with the first equation



{{{-y=4-2x}}}  Subtract {{{2x}}} from both sides



{{{-y=-2x+4}}} Rearrange the equation



{{{y=(-2x+4)/(-1)}}} Divide both sides by {{{-1}}}



{{{y=((-2)/(-1))x+(4)/(-1)}}} Break up the fraction



{{{y=2x-4}}} Reduce




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Since {{{y=2x-4}}}, we can now replace each {{{y}}} in the second equation with {{{2x-4}}} to solve for {{{x}}}




{{{2x-highlight((2x-4))=3}}} Plug in {{{y=2x-4}}} into the second equation. In other words, replace each {{{y}}} with {{{2x-4}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.




{{{2x-2x+4=3}}} Distribute the negative



{{{4=3}}} Combine like terms on the left side



Since this equation is <font size=4><b>NEVER</b></font> true for any x value, this means there are no solutions.



So the system is inconsistent.