Question

PLEASE HELP PLEASEEEEE

Part A: Explain why the x-coordinates of the points where the graphs of the equations y = 5−x and y = 2x + 1 intersect are the solutions of the equation 5−x = 2x + 1.
Part B: Make tables to find the solution to 5−x = 2x + 1. Take the integer values of x between −3 and 3.
Part C: How can you solve the equation 5−x = 2x + 1 grap hically?​

Answer 1

y = 5 - x → standard form y = -x + 5

y = 2x + 1

A: If two sides of an equation are set equal to each other, their value is a point of intersection because they will both equal a certain value.

B:

3 → 5 - 3 = 2(3) + 1 → 2 = 6 + 1 → 2 ≠ 7

2 → 5 - 2 = 2(2) + 1 → 3 = 4 + 1 → 3 ≠ 5

1 → 5 - 1 = 2(1) + 1 → 4 = 2 + 1 → 4 ≠ 3

0 → 5 - 0 = 2(0) + 1 → 5 = 0 + 1 → 5 ≠ 1

-1 → 5 - (-1) = 2(-1) + 1 → 6 = -2 + 1 → 6 ≠ -1

-2 → 5 - (-2) = 2(-2) + 1 → 7 = -4 + 1 → 7 ≠ -3

-3 → 5 - (-3) = 2(-3) + 1 → 8 = -6 + 1 → 8 ≠ -5

algebraically ↓

-x + 5 = 2x + 1 → 4 = 3x → x = 4/3

C: graphically

graph both equations and find point of intersection

Answer 2

Part A

We know these points are the solutions to
`5-x=2x+1` because by setting the two equations equal to each other, we find where they intersect, and those points are the solutions to the equation.

Part B

`\begin{array}c\cline{1-4}\bf x&\bf5-x&\bf2x+1&\bf5-x=2x+1\\\cline{1-4}-3&8&-5&8\\eq-5\\\cline{1-4}-2&7&-3&7\\eq-3\\\cline{1-4}-1&6&-1&6\\eq-1\\\cline{1-4}0&5&1&0\\eq1\\\cline{1-4}1&4&3&4\\eq3\\\cline{1-4}2&3&5&3\\eq5\\\cline{1-4}3&2&7&2\\eq7\\\cline{1-4}\end{array}`

Part B and a half

Since none of those values are correct solutions, let's solve this equation algebraically, even though the question doesn't ask for us to do so.

`\begin{aligned}5-x&=2x+1\\-x&=2x+1-5\\-x-2x&=1-5\\-3x&=-4\\x&=\boxed{(4)/(3)}\end{aligned}`

Part C

We can solve the equation graphically by graphing both
`y=5-x` and
`y=2x+1`, then finding where they meet. This is shown in the attachment below.