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Algebra 7 Online
OpenStudy (anonymous):

plz help Mary thought of two numbers. She doubled her first number and trebled the second, added these two together and got 5. Then she multiplied her first number by four and added this to her second number. This time she got a result of 15. Find the first two numbers.

OpenStudy (anonymous):

Okay, let's see if this works :P We'll call the first number x and the second y. 2x+3y=5 Then... 4x+y=15 To solve these you could put them onto a graph, by rewriting them. 3y=5-2x and y=15-4x Hopefully this should work! You should get 2 results for numbers on the x axis, which are your answers...

OpenStudy (anonymous):

okay thanks

OpenStudy (anonymous):

thats right!

OpenStudy (anonymous):

\[\text{Take it like this ..}\]\[\large\color{blue}{2x+3y=5}\]\[\large\color{green}{4x+y=15}\]\[\text{Now solve for x and y using substitution or any other method...}\]

OpenStudy (anonymous):

are the answers 4 and -1

OpenStudy (anonymous):

\[\text{You could do it like this..}\]\[4x+y=15\]\[x=\frac{15-y}{4}\]Now substitute this..\[3y=5-2x\]\[3y=5-2(\frac{15-y}{4})\]\[3y=5-(\frac{15-y}{2})\]\[6y=10-15+y\]\[5y=-5\]\[y=-1\]\[\text{So to find x}\]\[x=\frac{15-y}{4}\]\[x=\frac{15-(-1)}{4}\]\[x=\frac{16}{4}\]\[x=4\]

OpenStudy (anonymous):

YES

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