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​ALGEBRA 1 STATE TEST PRACTICE

Solutions to a linear / quadratic system of equations

10/4/2016

 
What are the solutions to the system of equations?
y = 3x - 2
​y = x²

x² = 3x - 2   {substituted x², in for y, into first equation}
x² - 3x + 2 = 0   {subtracted 3x and added 2 to each side}
(x - 2)(x - 1) = 0   {factored into two binomials}
x - 2 = 0   or   x - 1 = 0    {set each factor equal to 0}
x = 2   or  x = 1   {solved each equation for x}

If x = 2
y = x²   {second original equation}
y = 2²   {substituted 2, in for x, into y = x²}
y = 4   {evaluated exponent}
(2,4) is a solution

If x = 1
y = x²   {second original equation}
y = 1²   {substituted 1, in for x, into y = x²}
y = 1   {evaluated exponent}
(1,1) is  a solution

(2,4) and (1,1) are solutions to the system of equations.

When graphed, y = 3x - 2 is a straight line, and y = x² is a parabola.
The solutions are the points of intersection of the straight line and the parabola.
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