Answer :
To determine which equation could be Sandra's, we'll analyze the given equations and see if they have the same solutions as Tomas's equation, [tex]\( y = 3x + \frac{3}{4} \)[/tex].
Let's start by examining each of the given equations one at a time:
1. For the equation [tex]\(-6x + y = \frac{3}{2}\)[/tex]:
Rewrite Tomas's equation [tex]\( y = 3x + \frac{3}{4} \)[/tex] in such a way that we can substitute [tex]\( y \)[/tex]:
[tex]\[ y = 3x + \frac{3}{4} \][/tex]
Substitute [tex]\( y \)[/tex] in Sandra's equation:
[tex]\[ -6x + (3x + \frac{3}{4}) = \frac{3}{2} \][/tex]
Simplify:
[tex]\[ -6x + 3x + \frac{3}{4} = \frac{3}{2} \][/tex]
[tex]\[ -3x + \frac{3}{4} = \frac{3}{2} \][/tex]
To isolate [tex]\( x \)[/tex], move [tex]\(\frac{3}{4} \)[/tex] to the other side:
[tex]\[ -3x = \frac{3}{2} - \frac{3}{4} \][/tex]
Find a common denominator to combine the fractions:
[tex]\[ -3x = \frac{6}{4} - \frac{3}{4} \][/tex]
[tex]\[ -3x = \frac{3}{4} \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ x = -\frac{1}{4} \][/tex]
Substitute [tex]\( x = -\frac{1}{4} \)[/tex] back into Tomas's equation to find [tex]\( y \)[/tex]:
[tex]\[ y = 3(-\frac{1}{4}) + \frac{3}{4} \][/tex]
[tex]\[ y = -\frac{3}{4} + \frac{3}{4} \][/tex]
[tex]\[ y = 0 \][/tex]
Check if [tex]\((-\frac{1}{4}, 0)\)[/tex] satisfies [tex]\(-6x + y = \frac{3}{2}\)[/tex]:
[tex]\[ -6(-\frac{1}{4}) + 0 = \frac{3}{2} \][/tex]
[tex]\[ \frac{3}{2} = \frac{3}{2} \][/tex]
It holds true, so Sandra's equation [tex]\(-6x + y = \frac{3}{2}\)[/tex] has all the same solutions as Tomas's.
2. For the other equations, similar analysis shows:
[tex]\[ 6x + y = \frac{3}{2} \][/tex]
does not yield consistent solutions with Tomas's equation.
[tex]\[ -6x + 2y = \frac{3}{2} \][/tex]
also does not yield consistent solutions with Tomas's equation.
[tex]\[ 6x + 2y = \frac{3}{2} \][/tex]
lastly, also fails to have consistent solutions.
Hence, the correct equation that Sandra could have written, which has all the same solutions as Tomas's equation [tex]\(y = 3x + \frac{3}{4}\)[/tex], is:
[tex]\[ -6x + y = \frac{3}{2} \][/tex]
Therefore, the answer to which equation Sandra could have written is:
[tex]\[ \boxed{1} \][/tex]
Let's start by examining each of the given equations one at a time:
1. For the equation [tex]\(-6x + y = \frac{3}{2}\)[/tex]:
Rewrite Tomas's equation [tex]\( y = 3x + \frac{3}{4} \)[/tex] in such a way that we can substitute [tex]\( y \)[/tex]:
[tex]\[ y = 3x + \frac{3}{4} \][/tex]
Substitute [tex]\( y \)[/tex] in Sandra's equation:
[tex]\[ -6x + (3x + \frac{3}{4}) = \frac{3}{2} \][/tex]
Simplify:
[tex]\[ -6x + 3x + \frac{3}{4} = \frac{3}{2} \][/tex]
[tex]\[ -3x + \frac{3}{4} = \frac{3}{2} \][/tex]
To isolate [tex]\( x \)[/tex], move [tex]\(\frac{3}{4} \)[/tex] to the other side:
[tex]\[ -3x = \frac{3}{2} - \frac{3}{4} \][/tex]
Find a common denominator to combine the fractions:
[tex]\[ -3x = \frac{6}{4} - \frac{3}{4} \][/tex]
[tex]\[ -3x = \frac{3}{4} \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ x = -\frac{1}{4} \][/tex]
Substitute [tex]\( x = -\frac{1}{4} \)[/tex] back into Tomas's equation to find [tex]\( y \)[/tex]:
[tex]\[ y = 3(-\frac{1}{4}) + \frac{3}{4} \][/tex]
[tex]\[ y = -\frac{3}{4} + \frac{3}{4} \][/tex]
[tex]\[ y = 0 \][/tex]
Check if [tex]\((-\frac{1}{4}, 0)\)[/tex] satisfies [tex]\(-6x + y = \frac{3}{2}\)[/tex]:
[tex]\[ -6(-\frac{1}{4}) + 0 = \frac{3}{2} \][/tex]
[tex]\[ \frac{3}{2} = \frac{3}{2} \][/tex]
It holds true, so Sandra's equation [tex]\(-6x + y = \frac{3}{2}\)[/tex] has all the same solutions as Tomas's.
2. For the other equations, similar analysis shows:
[tex]\[ 6x + y = \frac{3}{2} \][/tex]
does not yield consistent solutions with Tomas's equation.
[tex]\[ -6x + 2y = \frac{3}{2} \][/tex]
also does not yield consistent solutions with Tomas's equation.
[tex]\[ 6x + 2y = \frac{3}{2} \][/tex]
lastly, also fails to have consistent solutions.
Hence, the correct equation that Sandra could have written, which has all the same solutions as Tomas's equation [tex]\(y = 3x + \frac{3}{4}\)[/tex], is:
[tex]\[ -6x + y = \frac{3}{2} \][/tex]
Therefore, the answer to which equation Sandra could have written is:
[tex]\[ \boxed{1} \][/tex]