Does the system have no solution?
1 answer:
Answer:
Step-by-step explanation:
Notice that in the second equation, the coefficient of is
times greater than in the first. Therefore, by making the coefficient of
in the second equation
times greater than in the first, it's impossible to only isolate one variable.
Therefore, when , there are no solutions to this system.
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2
Suggesting that you want this in standard form, in terms of quadratic equations, you would technically follow a process similar if not almost exactly like the 2 - step equation method with the exception of separating the (x)s and the equations to find x and then plug it in and what-not.
With that being said you would subtract 5 in (x+5) from said 5 in the second equation and -10 in the first equation in order to get 2x^2+7x-15, you would continue to do the same for the x by subtracting it from both ends making the 7x a 6x because there is a 1 at the beginning of each x if there is no number that is shown already. Which finally gives you the equation (y= 2x^2+6x-15)
Suggesting that you want this in standard form, in terms of quadratic equations, you would technically follow a process similar if not almost exactly like the 2 - step equation method with the exception of separating the (x)s and the equations to find x and then plug it in and what-not.
With that being said you would subtract 5 in (x+5) from said 5 in the second equation and -10 in the first equation in order to get 2x^2+7x-15, you would continue to do the same for the x by subtracting it from both ends making the 7x a 6x because there is a 1 at the beginning of each x if there is no number that is shown already. Which finally gives you the equation (y= 2x^2+6x-15)