Question
Use the substitution method to solve the system 2x+3y=4
2
𝑥
+
3
𝑦
=
4
, −4x−6y=−8
−
4
𝑥
−
6
𝑦
=
−
8
.(1 point)
Responses
infinite solutions
infinite solutions
(−1,2)
(
−
1
,
2
)
left parenthesis negative 1 comma 2 right parenthesis
(2−1)
(
2
−
1
)
left parenthesis 2 comma negative 1 right parenthesis
no solution
2
𝑥
+
3
𝑦
=
4
, −4x−6y=−8
−
4
𝑥
−
6
𝑦
=
−
8
.(1 point)
Responses
infinite solutions
infinite solutions
(−1,2)
(
−
1
,
2
)
left parenthesis negative 1 comma 2 right parenthesis
(2−1)
(
2
−
1
)
left parenthesis 2 comma negative 1 right parenthesis
no solution
Answers
Answered by
GPT-4o mini
To use the substitution method to solve the system of equations:
1. \(2x + 3y = 4\) (Equation 1)
2. \(-4x - 6y = -8\) (Equation 2)
First, let's simplify Equation 2:
We can divide the entire equation by -2:
\[
\frac{-4x}{-2} + \frac{-6y}{-2} = \frac{-8}{-2}
\]
This simplifies to:
\[
2x + 3y = 4
\]
Now we see that both equations are the same:
1. \(2x + 3y = 4\) (from Equation 1)
2. \(2x + 3y = 4\) (from the simplified Equation 2)
Since both equations are identical, they represent the same line in the coordinate plane. This means that there are infinitely many solutions along that line.
Thus, the solution to the system is:
**Infinite solutions.**
1. \(2x + 3y = 4\) (Equation 1)
2. \(-4x - 6y = -8\) (Equation 2)
First, let's simplify Equation 2:
We can divide the entire equation by -2:
\[
\frac{-4x}{-2} + \frac{-6y}{-2} = \frac{-8}{-2}
\]
This simplifies to:
\[
2x + 3y = 4
\]
Now we see that both equations are the same:
1. \(2x + 3y = 4\) (from Equation 1)
2. \(2x + 3y = 4\) (from the simplified Equation 2)
Since both equations are identical, they represent the same line in the coordinate plane. This means that there are infinitely many solutions along that line.
Thus, the solution to the system is:
**Infinite solutions.**
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