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6(y-4)=x-3

y=C
In the system of equations,
C is a constant. For which value of
C is
(x,y)=(-3,3) a solution?
Choose 1 answer:
(A) 3
(B) 21
(C) All real numbers
(D) None of the above

$6(y-4)=x-3$ $\newline$ $y=C$ $\newline$ In the system of equations, $C$ is a constant. For which value of $C$ is $(x, y)=(-3,3)$ a solution? $\newline$ Choose $1$ answer: $\newline$ (A) $3$ $\newline$ (B) $21$ $\newline$ (C) All real numbers $\newline$ (D) None of the above

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Find the number of solutions to a system of equations

Full solution

Q.

6(y-4)=x-3

\newline

y=C

\newline

In the system of equations,

C

is a constant. For which value of

C

is

(x, y)=(-3,3)

a solution?

\newline

Choose

1

answer:

\newline

(A)

3

\newline

(B)

21

\newline

(C) All real numbers

\newline

(D) None of the above

Substitute $y = C$ : Substitute $y = C$ into the equation $6(y-4) = x-3$ . Since $y = C$ , the equation becomes $6(C-4) = x-3$ .
Plug in values: Plug in the values $x = -3$ and $y = 3$ from the solution point $(-3,3)$ into the equation. This gives us $6(3-4) = -3 - 3$ .
Simplify left side: Simplify the left side: $6(-1) = -6$ . Now, the equation is $-6 = -6$ .
Check against original substitution: Since the equation $-6 = -6$ holds true, the value of $C$ that makes $(x,y) = (-3,3)$ a solution must be checked against the original substitution. We substituted $y = C$ , and since $y = 3$ in the solution point, $C$ must be $3$ .

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