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How do you substitute equations y5x 5 y15x-1? - Answers

To substitute the equations ( y = 5x ) and ( y = 15x - 1 ), set them equal to each other: ( 5x = 15x - 1 ). Then, solve for ( x ) by rearranging the equation: ( 1 = 15x - 5x ) or ( 10x = 1 ), giving ( x = \frac{1}{10} ). Substitute ( x ) back into either equation to find ( y ); using ( y = 5x ) gives ( y = 5 \times \frac{1}{10} = \frac{1}{2} ). Thus, the point of intersection is ( \left( \frac{1}{10}, \frac{1}{2} \right) ).



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How do you substitute equations y5x 5 y15x-1? - Answers

https://math.answers.com/math-and-arithmetic/How_do_you_substitute_equations_y5x_5_y15x-1

To substitute the equations ( y = 5x ) and ( y = 15x - 1 ), set them equal to each other: ( 5x = 15x - 1 ). Then, solve for ( x ) by rearranging the equation: ( 1 = 15x - 5x ) or ( 10x = 1 ), giving ( x = \frac{1}{10} ). Substitute ( x ) back into either equation to find ( y ); using ( y = 5x ) gives ( y = 5 \times \frac{1}{10} = \frac{1}{2} ). Thus, the point of intersection is ( \left( \frac{1}{10}, \frac{1}{2} \right) ).



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https://math.answers.com/math-and-arithmetic/How_do_you_substitute_equations_y5x_5_y15x-1

How do you substitute equations y5x 5 y15x-1? - Answers

To substitute the equations ( y = 5x ) and ( y = 15x - 1 ), set them equal to each other: ( 5x = 15x - 1 ). Then, solve for ( x ) by rearranging the equation: ( 1 = 15x - 5x ) or ( 10x = 1 ), giving ( x = \frac{1}{10} ). Substitute ( x ) back into either equation to find ( y ); using ( y = 5x ) gives ( y = 5 \times \frac{1}{10} = \frac{1}{2} ). Thus, the point of intersection is ( \left( \frac{1}{10}, \frac{1}{2} \right) ).

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      To substitute the equations ( y = 5x ) and ( y = 15x - 1 ), set them equal to each other: ( 5x = 15x - 1 ). Then, solve for ( x ) by rearranging the equation: ( 1 = 15x - 5x ) or ( 10x = 1 ), giving ( x = \frac{1}{10} ). Substitute ( x ) back into either equation to find ( y ); using ( y = 5x ) gives ( y = 5 \times \frac{1}{10} = \frac{1}{2} ). Thus, the point of intersection is ( \left( \frac{1}{10}, \frac{1}{2} \right) ).

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