Question 180144
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Given the system:


*[tex \Large \text {          }\math y=5x+5]
*[tex \Large \text {          }\math y=15x-1]


*[tex \Large \text {          }\math 5x+5=15x-1] is a good start.  Now all you need to do is solve for <i>x</i> and then substitute that value back into either one of the original equations to compute the value of <i>y</i>.



*[tex \Large \text {          }\math 5x+5=15x-1]


Add *[tex \Large -5x + 1] to both sides:


*[tex \Large \text {          }\math 6=10x]


Multiply both sides by *[tex \Large {1 \over 10}]


*[tex \Large \text {          }\math x = {6 \over 10} = {3 \over 5}]


Substitute in the first of the original equations:


*[tex \Large \text {          }\math y=5({3 \over 5})+5 = 3 + 5 = 8]


So your solution set is *[tex \Large \left \{({3 \over 5},8)\right}]


Check your answer by substituting for <i>x</i> and <i>y</i> in the 2nd equation of your original system:


Is *[tex \Large 8=15({3 \over 5})-1] a true statement?


*[tex \Large 8=15({3 \over 5})-1 \rightarrow 8 = 3(3) - 1 \rightarrow 8 = 9 -1]


True, answer checks.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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