Question 135442
Start with the given system

{{{y=12x-19}}}


{{{y=x^2+5x-7}}}




{{{12x-19=x^2+5x-7}}} Set the two equations equal to one another




{{{0=x^2+5x-7-12x+19}}}  Subtract 12x from both sides.  Add 19 to both sides. 




{{{0=x^2-7x+12}}}  Combine like terms



{{{(x-4)(x-3)=0}}} Factor the left side (note: if you need help with factoring, check out this <a href=http://www.algebra.com/algebra/homework/playground/change-this-name4450.solver>solver</a>)




Now set each factor equal to zero:

{{{x-4=0}}} or  {{{x-3=0}}} 


{{{x=4}}} or  {{{x=3}}}    Now solve for x in each case



So our x values are


 {{{x=4}}} or  {{{x=3}}} 




{{{y=12x-19}}} Start with the first equation


{{{y=12(4)-19}}} Plug in {{{x=4}}}


{{{y=29}}} Simplify



So the first ordered pair is (4,29)




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{{{y=12x-19}}} Start with the first equation


{{{y=12(3)-19}}} Plug in {{{x=3}}}


{{{y=17}}} Simplify



So the second ordered pair is (3,17)





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Answer:


So the solutions are 

(4,29) or (3,17)