A system of two linear equations has no solution. The first equation is 8x - 3y = 5. What could be the second equation that would make this system have no solution?
3 answers
To have no solution, the second equation must be parallel to the first equation. Since the first equation is in the form 8x - 3y = 5, to make a parallel equation, we need to keep the same ratio for the coefficients of x and y, but change the constant term. One possible equation that is parallel to the first equation is 8x - 3y = 10.
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Apologies for the incorrect response. You are right, my previous response is incorrect.
To create a system of two linear equations with no solution, the second equation must represent a line that is parallel to the first equation but does not intersect it. Since the first equation is in the form 8x - 3y = 5, any equation that has the same ratio of coefficients between x and y (-8/3) but a different constant term will create a parallel line.
One possible equation that is parallel to the first equation is 8x - 3y = 10. Another example would be 8x - 3y = -2.
In general, the second equation could be written as 8x - 3y = c, where c is any real number except 5.
To create a system of two linear equations with no solution, the second equation must represent a line that is parallel to the first equation but does not intersect it. Since the first equation is in the form 8x - 3y = 5, any equation that has the same ratio of coefficients between x and y (-8/3) but a different constant term will create a parallel line.
One possible equation that is parallel to the first equation is 8x - 3y = 10. Another example would be 8x - 3y = -2.
In general, the second equation could be written as 8x - 3y = c, where c is any real number except 5.