How do I solve this using any algebraic method? 3x-2y=10 5x+3y=15 2x-4y=-6 -x+2y=3
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solve for x on one equation and substitute it in the other
Originally Posted by Vuong How do I solve this using any algebraic method? 3(3x-2y=10) 2(5x+3y=15) 2x-4y=-6 2(-x+2y=3) solve using elimination ... distribute, add the equations and solve for the remaining variable. back substitute to find the other variable.
From the second equation you know that : Thus And so, . Substituting back this result into the first equation gives : Now substitute the value of found into any of the equations to retrieve . Can you do it for the second system now ?
Originally Posted by Vuong 3x-2y=10 5x+3y=15 You can also use the elimination method make ...(1) ...(2) To eliminate x take (1) and (2) gives ...(3) ...(4) Now taking (4) form (3) gives Can you take it from here?
Originally Posted by Bacterius From the second equation you know that : Thus And so, . Substituting back this result into the first equation gives : Now substitute the value of found into any of the equations to retrieve . Can you do it for the second system now ? How did you get the 9 for ?
Originally Posted by pickslides You can also use the elimination method make ...(1) ...(2) To eliminate x take (1) and (2) gives ...(3) ...(4) Now taking (4) form (3) gives Can you take it from here? y = -3.8 then plug it in?
Originally Posted by Vuong How did you get the 9 for ? Comes from the substitution of .
Originally Posted by Vuong y = -3.8 then plug it in? To either (1),(2),(3) or (4) to find x.
Originally Posted by pickslides To either (1),(2),(3) or (4) to find x. Thanks, I got 5.86
That's not right, you should find something around
Originally Posted by Bacterius That's not right, you should find something around I plugged -3.8 into 3x-2y=10 x_x
Originally Posted by Vuong y = -3.8 then plug it in? Are you sure this is correct?
Originally Posted by pickslides Are you sure this is correct? So I plug -0.263 into one of them?
Originally Posted by Vuong So I plug -0.263 into one of them?
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