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y = x^{2} − 4x + 3

y = x − 1

If (x, y) is a solution to the system of equations above, what is one possible value of the product of x and y ?

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y = x^{2} − 4x + 3 ----------- eqn(1)

y = x − 1 ------------- eqn(2)

equate eqn(1) and eqn(2),

x^{2} − 4x + 3 = x − 1

collect like term

x^{2} − 4x - x + 3 + 1 = 0

x^{2} − 5x + 4 = 0 ---------- eqn(1)

The above is a quadratic equation and can be simplified either by factorization, completing the square or quadratic formula method of solving quadratic equation.

Using factorization method,

from eqn(1),

sum of roots = -5

product of roots = 1 x 4 = 4

Find two numbers whose sum and product is -5 and 4 respectively.

Let the two numbers be -1 and -4

Replace the sum of roots (-5) in eqn(1) with these two numbers.

eqn(1) becomes:

x^{2} − 1x - 4x + 4 = 0

x^{2} − x - 4x + 4 = 0

Next, group into two expressions

(x^{2} − x) - (4x - 4) = 0

factories both expressions

x(x − 1) - 4(x - 1) = 0

Pick and group both terms outside each bracket and inside one of the two brackets. That is,

(x - 4)(x - 1) = 0

x - 4 = 0 and x - 1 = 0

x = 4 and x = 1

substitute each value of x into eqn(2)

if x = 4,

y = 4 − 1

y = 3

Thus, (4, 3)

if x = 1

y = 1 − 1

y = 0

Thus, (1, 0)

Hence, the solutions, (x, y) are:

(4, 3) and (1, 0)

The product of each solution is:

4 * 3 = 12

and (1 * 0) = 0

Therefore, the product of the solutions are: 0, 12. But since the question asked for only one possible value, either 0 or 12 is correct

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