Question 975762
y = x² + 3 
y = x + 5


x + 5 = x² + 3 {substituted (x + 5), in for y, into top equation}
x² - x - 2 = 0 {subtracted x and 2 from 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}


y = x = 5 {bottom original equation}
y = 2 + 5 or y = -1 + 5 {substituted 2 and -1, in for x, into bottom original equation}
y = 7 or y = 4 {solved each equation for y}


x = 2 and y = 7
or 
x = -1 and y = 4
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