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Mathematics 7 Online
OpenStudy (anonymous):

Solve this system of equations. x+2y=4 2x+y=10

OpenStudy (anonymous):

take one of the equations and solve for either y or x

OpenStudy (anonymous):

then use that to substitute in for the other equation

OpenStudy (anonymous):

i got (6, -2) . is that right ?

OpenStudy (anonymous):

hold on let me check since i haven't done the calculation

OpenStudy (anonymous):

k

OpenStudy (anonymous):

check to see whether the (x,y) you got satisfies both equations (by subbing them into both and making sure that the LHS equals the RHS). If so, then yes you are right.

OpenStudy (anonymous):

or you can do that lol

OpenStudy (anonymous):

(6,-2) So does: x+2y=4 ---> (6) +(2)(-2) = 4 ? no. 2x+y=10 ---> (2)(6) + (-2) = 10 ? yes Somethings wrong.

OpenStudy (anonymous):

hmmmm.

OpenStudy (anonymous):

im confused.

OpenStudy (anonymous):

\[x+2y=4\to x= 4-2y\] \[2x+y=10 \\ 2(4-2y) +y =10\\ 8-4y+y=10\\-3y=2\\y=-\frac{2}{3}\] Can you find x from here?

OpenStudy (anonymous):

yes,

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