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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. Let's consider the first equation
When you examine the first equation, it is factorable by the product of two binomials. Hence, the factors are
Equate each factor to zero and solve for y in terms of x.
For (2x + 3y),
Substitute the value of y to the second given equation, we have
If you will choose positive sign, then the value of y is
If you will choose negative sign, then the value of y is
For (x - y),
Substitute the value of y to the second given equation, we have
If you will choose positive sign, then the value of y is
If you will choose negative sign, then the value of y is
Therefore, the solutions of the two equations are:
