Question 308972



Start with the given system of equations:


{{{system(2x+3y=4,3x+4y=5)}}}




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+3y=4}}} Start with the first equation



{{{3y=4-2x}}}  Subtract {{{2x}}} from both sides



{{{3y=-2x+4}}} Rearrange the equation



{{{y=(-2x+4)/(3)}}} Divide both sides by {{{3}}}



{{{y=((-2)/(3))x+(4)/(3)}}} Break up the fraction



{{{y=(-2/3)x+4/3}}} Reduce




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


Since {{{y=(-2/3)x+4/3}}}, we can now replace each {{{y}}} in the second equation with {{{(-2/3)x+4/3}}} to solve for {{{x}}}




{{{3x+4highlight(((-2/3)x+4/3))=5}}} Plug in {{{y=(-2/3)x+4/3}}} into the second equation. In other words, replace each {{{y}}} with {{{(-2/3)x+4/3}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.




{{{3x+(4)(-2/3)x+(4)(4/3)=5}}} Distribute {{{4}}} to {{{(-2/3)x+4/3}}}



{{{3x-(8/3)x+16/3=5}}} Multiply



{{{(3)(3x-(8/3)x+16/3)=(3)(5)}}} Multiply both sides by the LCM of 3. This will eliminate the fractions  (note: if you need help with finding the LCM, check out this <a href=http://www.algebra.com/algebra/homework/divisibility/least-common-multiple.solver>solver</a>)




{{{9x-8x+16=15}}} Distribute and multiply the LCM to each side




{{{x+16=15}}} Combine like terms on the left side



{{{x=15-16}}}Subtract 16 from both sides



{{{x=-1}}} Combine like terms on the right side






-----------------First Answer------------------------------



So the first part of our answer is: {{{x=-1}}}










Since we know that {{{x=-1}}} we can plug it into the equation {{{y=(-2/3)x+4/3}}} (remember we previously solved for {{{y}}} in the first equation).




{{{y=(-2/3)x+4/3}}} Start with the equation where {{{y}}} was previously isolated.



{{{y=(-2/3)(-1)+4/3}}} Plug in {{{x=-1}}}



{{{y=2/3+4/3}}} Multiply



{{{y=2}}} Combine like terms and reduce.  (note: if you need help with fractions, check out this <a href="http://www.algebra.com/algebra/homework/NumericFractions/fractions-solver.solver">solver</a>)




-----------------Second Answer------------------------------



So the second part of our answer is: {{{y=2}}}










-----------------Summary------------------------------


So our answers are:


{{{x=-1}}} and {{{y=2}}}


which form the point *[Tex \LARGE \left(-1,2\right)] 









Now let's graph the two equations (if you need help with graphing, check out this <a href=http://www.algebra.com/algebra/homework/Linear-equations/graphing-linear-equations.solver>solver</a>)



From the graph, we can see that the two equations intersect at *[Tex \LARGE \left(-1,2\right)]. This visually verifies our answer.





{{{
drawing(500, 500, -10,10,-10,10,
  graph(500, 500, -10,10,-10,10, (4-2*x)/(3), (5-3*x)/(4) ),
  blue(circle(-1,2,0.1)),
  blue(circle(-1,2,0.12)),
  blue(circle(-1,2,0.15))
)
}}} graph of {{{2x+3y=4}}} (red) and {{{3x+4y=5}}} (green)  and the intersection of the lines (blue circle).