Question #46dc6

1 Answer
SCooke
Dec 4, 2017

Solutions:
#x = -1, y = 3#
#x = -1/3, y = 3 2/3#

Explanation:

We can find the solution (common points of each equation) algebraically or graphically. To make a graph, simply make a table with 'x', 'y1' (from expression1), 'y2' (from expression2) and then plot them on graph paper.
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Algebraically, we can use the first equation to set y = x + 4 and then substitute it into the second equation.
#2x^2 + x(x + 4) = -1# ; #2x^2 + x^2 + 4x + 1 = 0#
#3x^2 + 4x + 1 = 0# This can be solved with the quadratic formula.

For ax2 + bx + c = 0, the values of x which are the solutions of the equation are given by:
#x = ​(−b±√[​b​^2​​−4ac])/(2a)#
In this case, a = 3, b = 4 and c = 1
#x = ​(−4 ± sqrt[(4(^​2)​ ​− 4*3*1]))/(2*3)#
#x = (​−4 ± sqrt[16​ ​− 12])/(6)#
#x = ​(−4 ± 2)/6#; #x = -6/6 and -2/6#
We can see that the algebraic solution may be more precise than we could obtain visually from a graph.

CHECK: x = -1 , y = 3
#2(-1)^2 + (-1) xx (3) = -1# ; #2 – 3 = -1# ; #-1 = -1# CORRECT

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