Which ordered pair is a solution of the equation? y-4=7(x-6) Choose 1 answer: (A) Only(5,4) (B) Only (6,5) (c) Both (5,4) and (6,5) (D) Neither

Understanding the equation: Understand the equation and the form it is in. The equation given is y - 4 = 7(x - 6). This is a linear equation in a non-standard form. To find the solution, we need to plug in the x and y values from the given ordered pairs to see if they satisfy the equation.Testing the ordered pair (5, 4): Test the ordered pair (5, 4). Substitute x = 5 and y = 4 into the equation and check if the left side equals the right side. 4 - 4 = 7(5 - 6) 0 = 7(-1) 0 = -7 This is not true, so (5, 4) is not a solution to the equation.Testing the ordered pair (6, 5): Test the ordered pair (6, 5). Substitute x = 6 and y = 5 into the equation and check if the left side equals the right side. 5 - 4 = 7(6 - 6) 1 = 7(0) 1 = 0 This is also not true, so (6, 5) is not a solution to the equation.

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Algebra 1

Does (x, y) satisfy the linear equation?

Full solution

Q. Which ordered pair is a solution of the equation?

\newline

y-4=7(x-6)

\newline

Choose

1

answer:

\newline

(A) Only

(5,4)

\newline

(B) Only

(6,5)

\newline

(5,4)

and

(6,5)

\newline

(D) Neither

Understanding the equation: Understand the equation and the form it is in. $\newline$ The equation given is $y - 4 = 7(x - 6)$ . This is a linear equation in a non-standard form. To find the solution, we need to plug in the $x$ and $y$ values from the given ordered pairs to see if they satisfy the equation.
Testing the ordered pair $(5, 4)$ : Test the ordered pair $(5, 4)$ . $\newline$ Substitute $x = 5$ and $y = 4$ into the equation and check if the left side equals the right side. $\newline$ $4 - 4 = 7(5 - 6)$ $\newline$ $0 = 7(-1)$ $\newline$ $0 = -7$ $\newline$ This is not true, so $(5, 4)$ is not a solution to the equation.
Testing the ordered pair $(6, 5)$ : Test the ordered pair $(6, 5)$ . $\newline$ Substitute $x = 6$ and $y = 5$ into the equation and check if the left side equals the right side. $\newline$ $5 - 4 = 7(6 - 6)$ $\newline$ $1 = 7(0)$ $\newline$ $1 = 0$ $\newline$ This is also not true, so $(6, 5)$ is not a solution to the equation.