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Solve the following systems by substitution:
Solution:
Consider the given equations above
Did you notice that both equations have xy term aside from the second degree of x and y? Well, there's a special procedure in which we can solve for the value of x and y. We need to do the trial and error like factoring of equations (if there are factors), adding, subtracting, and even substitution.
If you add the two given equations, the left side of the resulting equation becomes factorable by perfect trinomial square.
Take the square root on both sides of the equation, we have
If you will choose positive sign, then the value of y in terms of x is
Substitute the value of y to the second given equation, we have
After equating each factor to zero, then the values of x are 3 and - 2.
If x = 3, then the value of y is
If x = - 2, then the value of y is
If you will choose negative sign, then the value of y in terms of x is
Substitute the value of y to the second given equation, we have
After equating each factor to zero, then the values of x are 2 and - 3.
If x = 2, then the value of y is
If x = - 3, then the value of y is
Therefore, the solutions of the two equations are:
