Question

name: block 4 algebra i sped teacher resource room math coach - ms. almonte u2. l10 homework due 10/9 1. here is a graph of the equation 2y - x = 1. a. are the points (0, -\frac{1}{2}) and (-7, -3) solutions to the equation? explain or show how you know.

Explanation:

Step1: Recall the definition of a solution to an equation

A point $(x,y)$ is a solution to the equation if it satisfies the equation when we substitute the $x$ - and $y$ - values into it. The given equation is $2y - x=1$.

Step2: Check the point $(0,-\frac{1}{2})$

Substitute $x = 0$ and $y=-\frac{1}{2}$ into the equation:
\[

$$\begin{align*} 2y - x&=2\times(-\frac{1}{2})-0\\ &=- 1-0\\ &=-1 eq1 \end{align*}$$

\]
So, $(0,-\frac{1}{2})$ is not a solution.

Step3: Check the point $(-7,-3)$

Substitute $x=-7$ and $y = - 3$ into the equation:
\[

$$\begin{align*} 2y - x&=2\times(-3)-(-7)\\ &=-6 + 7\\ &=1 \end{align*}$$

\]
So, $(-7,-3)$ is a solution.

Answer:

a. The point $(0,-\frac{1}{2})$ is not a solution to the equation $2y - x = 1$ because when $x = 0$ and $y=-\frac{1}{2}$, $2y - x=-1
eq1$. The point $(-7,-3)$ is a solution to the equation $2y - x = 1$ because when $x=-7$ and $y=-3$, $2y - x=-6 + 7=1$.