Question

Question #46dc6

Answer 1

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.

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