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How do you solve algebraically the system of equations? - Answers

You can solve a system of equations using a few methods: elemination and subsitution. In the elemination method you would multiply each of your equations by a factor that would cause one of the variables to be eleminated. ex: If you had 2x-3y=5 and 3x+6y=4 you would multiply the first equation by a factor of 2 (distributing the 2 to each term), giving you a new equation of 4x-6y=10. Now your y terms would cancel. You line up the equations like: 3x+6y=4 4x-6y=10 you simply cancel the y terms and add the others giving you: 7x=14 you then solve for x (in this case divide by 7) x=2 now you plug in x=2 into one of the first equations and solve 3x+6y=4 3(2)+6y=4 plug in x 6+6y=4 multiply 6y=-2 add like terms y=-1/3 solve for y To solve using the subsitution method you would take an equation and solve for one variable then plug that into the other equation. ex. (using the same equations as before) 2x-3y=5 2x=3y+5 x=(3/2)y+(5/2) 3x+6y=4 3((3/2)y+(5/2))+6y=4 plug in for x (9/2)y+(15/2)+6y=4 distribute the 3 (21/2)y=(-7/2) add like terms y=(-1/3) solve for y you would then plug in y and solve for x 2x-3(-1/3)=5 2x+1=5 2x=4 x=2 Hope this helps!



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How do you solve algebraically the system of equations? - Answers

https://math.answers.com/algebra/How_do_you_solve_algebraically_the_system_of_equations

You can solve a system of equations using a few methods: elemination and subsitution. In the elemination method you would multiply each of your equations by a factor that would cause one of the variables to be eleminated. ex: If you had 2x-3y=5 and 3x+6y=4 you would multiply the first equation by a factor of 2 (distributing the 2 to each term), giving you a new equation of 4x-6y=10. Now your y terms would cancel. You line up the equations like: 3x+6y=4 4x-6y=10 you simply cancel the y terms and add the others giving you: 7x=14 you then solve for x (in this case divide by 7) x=2 now you plug in x=2 into one of the first equations and solve 3x+6y=4 3(2)+6y=4 plug in x 6+6y=4 multiply 6y=-2 add like terms y=-1/3 solve for y To solve using the subsitution method you would take an equation and solve for one variable then plug that into the other equation. ex. (using the same equations as before) 2x-3y=5 2x=3y+5 x=(3/2)y+(5/2) 3x+6y=4 3((3/2)y+(5/2))+6y=4 plug in for x (9/2)y+(15/2)+6y=4 distribute the 3 (21/2)y=(-7/2) add like terms y=(-1/3) solve for y you would then plug in y and solve for x 2x-3(-1/3)=5 2x+1=5 2x=4 x=2 Hope this helps!



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https://math.answers.com/algebra/How_do_you_solve_algebraically_the_system_of_equations

How do you solve algebraically the system of equations? - Answers

You can solve a system of equations using a few methods: elemination and subsitution. In the elemination method you would multiply each of your equations by a factor that would cause one of the variables to be eleminated. ex: If you had 2x-3y=5 and 3x+6y=4 you would multiply the first equation by a factor of 2 (distributing the 2 to each term), giving you a new equation of 4x-6y=10. Now your y terms would cancel. You line up the equations like: 3x+6y=4 4x-6y=10 you simply cancel the y terms and add the others giving you: 7x=14 you then solve for x (in this case divide by 7) x=2 now you plug in x=2 into one of the first equations and solve 3x+6y=4 3(2)+6y=4 plug in x 6+6y=4 multiply 6y=-2 add like terms y=-1/3 solve for y To solve using the subsitution method you would take an equation and solve for one variable then plug that into the other equation. ex. (using the same equations as before) 2x-3y=5 2x=3y+5 x=(3/2)y+(5/2) 3x+6y=4 3((3/2)y+(5/2))+6y=4 plug in for x (9/2)y+(15/2)+6y=4 distribute the 3 (21/2)y=(-7/2) add like terms y=(-1/3) solve for y you would then plug in y and solve for x 2x-3(-1/3)=5 2x+1=5 2x=4 x=2 Hope this helps!

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      You can solve a system of equations using a few methods: elemination and subsitution. In the elemination method you would multiply each of your equations by a factor that would cause one of the variables to be eleminated. ex: If you had 2x-3y=5 and 3x+6y=4 you would multiply the first equation by a factor of 2 (distributing the 2 to each term), giving you a new equation of 4x-6y=10. Now your y terms would cancel. You line up the equations like: 3x+6y=4 4x-6y=10 you simply cancel the y terms and add the others giving you: 7x=14 you then solve for x (in this case divide by 7) x=2 now you plug in x=2 into one of the first equations and solve 3x+6y=4 3(2)+6y=4 plug in x 6+6y=4 multiply 6y=-2 add like terms y=-1/3 solve for y To solve using the subsitution method you would take an equation and solve for one variable then plug that into the other equation. ex. (using the same equations as before) 2x-3y=5 2x=3y+5 x=(3/2)y+(5/2) 3x+6y=4 3((3/2)y+(5/2))+6y=4 plug in for x (9/2)y+(15/2)+6y=4 distribute the 3 (21/2)y=(-7/2) add like terms y=(-1/3) solve for y you would then plug in y and solve for x 2x-3(-1/3)=5 2x+1=5 2x=4 x=2 Hope this helps!
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