# Question 46dc6

Dec 4, 2017

Solutions:
$x = - 1 , y = 3$
$x = - \frac{1}{3} , y = 3 \frac{2}{3}$

#### 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.

Algebraically, we can use the first equation to set y = x + 4 and then substitute it into the second equation.
$2 {x}^{2} + x \left(x + 4\right) = - 1$ ; $2 {x}^{2} + {x}^{2} + 4 x + 1 = 0$
$3 {x}^{2} + 4 x + 1 = 0$ 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; $x = - \frac{6}{6} \mathmr{and} - \frac{2}{6}$
We can see that the algebraic solution may be more precise than we could obtain visually from a graph.

CHECK: x = -1 , y = 3
$2 {\left(- 1\right)}^{2} + \left(- 1\right) \times \left(3\right) = - 1$ ; 2 – 3 = -1# ; $- 1 = - 1$ CORRECT