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A system of equations is shown below: -3x + 7y = -16 -9x + 5y = 16 Part A: Create an equivalent system of equations by replacing one equation with the sum of that equation and a multiple of the other. Show the steps to do this. Part B: Show that the equivalent system has the same solution as the original system of equations.
@triciaal hey can you help me please
so you need to solve by elimination.
OK so how do we do that.
the objective is to pick a variable and make the coefficient equal and opposite so that when you add the equations the sum of that variable is zero as in eliminated. you now have one equation with one variable. solve. use this value in either of the originals to get the value of the other.
so wait if this is elimination we would add them
i will walk you through this time
to get the x then use the x to get the y
pick a variable
so to make the coefficients equal and opposite multiply the first eqn by - 3
ok so that would look something like this 9x+7y=48
am i right?
do we subtract -10 with 64?
no you forgot to multiply the 7y
slow down and follow
now that you have y substitute in either of the originals and calculate x
wait what was our y
it is always a good practice to check the solution in the other original because it is very very easy to make arithmetic errors.
indeed I make them all the time xD
so what was your solution?
for what... what I said earlier?
oh wait so we were supposed to multiply the -3 *7
you forgot your y
oh yea 9x-21y=48
no wait im supposed to divide sorry
now find x
-3x + 7(-4) = -16
So we would do +3x
is-int it multiplication
go ahead and +3x if that is easier for you the idea is to isolate the variable terms to one side and the constants to the other then divide by the coefficient of the variable
both of them are -4 right?
solution (-4, -4)
So how would i do this ... Create an equivalent system of equations by replacing one equation with the sum of that equation and a multiple of the other. Show the steps to do this
review everything from the beginning
ok ... I dont see it
like I see what your talking about but I dont see how that is an equivalent system
@triciaal sorry to bother you more but I dont understand
where did you stop?
did you review? we have a multiple of equation 1 because we multiplied by -3 we added this to the 2nd eqn equivalent system created sum means to add
show it has the same result the solution works in each of the originals
oh yes I did and thanks so much for helping me