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How do you solve systems of equations using substitution in pre-algebra? - Answers

cant explain it but i can give you an example: equation 1: 14x+2y=20 equation 2: 5x-y=8 First step is to get on of the varibles to cancel out so leave the first eqation alone put change the second in this problem so: 5x-y=8 becomes 2(5x-y=8) which equals 10x-2y=16 the next step is to subtract them like this: 14x+2y=20 10x-2y=16 the +2y and the -2y cancel out so the answer then would be 4x=4 which means 1=x then you plug the answer 1 in for x into on of the equations to find y so: 5(1)-y=8 = 5-y=8 =-y=3 =y=-3 so the system answer is (1,-3)



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How do you solve systems of equations using substitution in pre-algebra? - Answers

https://math.answers.com/algebra/How_do_you_solve_systems_of_equations_using_substitution_in_pre-algebra

cant explain it but i can give you an example: equation 1: 14x+2y=20 equation 2: 5x-y=8 First step is to get on of the varibles to cancel out so leave the first eqation alone put change the second in this problem so: 5x-y=8 becomes 2(5x-y=8) which equals 10x-2y=16 the next step is to subtract them like this: 14x+2y=20 10x-2y=16 the +2y and the -2y cancel out so the answer then would be 4x=4 which means 1=x then you plug the answer 1 in for x into on of the equations to find y so: 5(1)-y=8 = 5-y=8 =-y=3 =y=-3 so the system answer is (1,-3)



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https://math.answers.com/algebra/How_do_you_solve_systems_of_equations_using_substitution_in_pre-algebra

How do you solve systems of equations using substitution in pre-algebra? - Answers

cant explain it but i can give you an example: equation 1: 14x+2y=20 equation 2: 5x-y=8 First step is to get on of the varibles to cancel out so leave the first eqation alone put change the second in this problem so: 5x-y=8 becomes 2(5x-y=8) which equals 10x-2y=16 the next step is to subtract them like this: 14x+2y=20 10x-2y=16 the +2y and the -2y cancel out so the answer then would be 4x=4 which means 1=x then you plug the answer 1 in for x into on of the equations to find y so: 5(1)-y=8 = 5-y=8 =-y=3 =y=-3 so the system answer is (1,-3)

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      cant explain it but i can give you an example: equation 1: 14x+2y=20 equation 2: 5x-y=8 First step is to get on of the varibles to cancel out so leave the first eqation alone put change the second in this problem so: 5x-y=8 becomes 2(5x-y=8) which equals 10x-2y=16 the next step is to subtract them like this: 14x+2y=20 10x-2y=16 the +2y and the -2y cancel out so the answer then would be 4x=4 which means 1=x then you plug the answer 1 in for x into on of the equations to find y so: 5(1)-y=8 = 5-y=8 =-y=3 =y=-3 so the system answer is (1,-3)

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