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https://math.answers.com/math-and-arithmetic/Does_2y_x_4_and_y_12_3x_have_a_common_solution

Does 2y x 4 and y 12 3x have a common solution? - Answers

Assuming that the two equations are: 2y = x + 4 .................................. 1 y = 12 + 3x ............................. 2 multiplying eqn 2 by two and subtracting it from eqn 1 results in: 0 = 5x + 20, then x = -4 Substituting in eqn 2 with x = -4 results in y = 0 then the common intersection point is (-4, 0)



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Does 2y x 4 and y 12 3x have a common solution? - Answers

https://math.answers.com/math-and-arithmetic/Does_2y_x_4_and_y_12_3x_have_a_common_solution

Assuming that the two equations are: 2y = x + 4 .................................. 1 y = 12 + 3x ............................. 2 multiplying eqn 2 by two and subtracting it from eqn 1 results in: 0 = 5x + 20, then x = -4 Substituting in eqn 2 with x = -4 results in y = 0 then the common intersection point is (-4, 0)



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https://math.answers.com/math-and-arithmetic/Does_2y_x_4_and_y_12_3x_have_a_common_solution

Does 2y x 4 and y 12 3x have a common solution? - Answers

Assuming that the two equations are: 2y = x + 4 .................................. 1 y = 12 + 3x ............................. 2 multiplying eqn 2 by two and subtracting it from eqn 1 results in: 0 = 5x + 20, then x = -4 Substituting in eqn 2 with x = -4 results in y = 0 then the common intersection point is (-4, 0)

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      Assuming that the two equations are: 2y = x + 4 .................................. 1 y = 12 + 3x ............................. 2 multiplying eqn 2 by two and subtracting it from eqn 1 results in: 0 = 5x + 20, then x = -4 Substituting in eqn 2 with x = -4 results in y = 0 then the common intersection point is (-4, 0)

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