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Math  /  Algebra

Question11) x+3y=1−3x−3y=−15x=0,y=13\begin{array}{l}x+3 y=1 \\ -3 x-3 y=-15 \\ x=0, y=\frac{1}{3}\end{array}

Studdy Solution

STEP 1

1. We have a system of linear equations: $ \begin{align*} x + 3y &= 1 \\ -3x - 3y &= -15 \end{align*} \]
2. We are given a potential solution: \( x = 0 \) and \( y = \frac{1}{3} \).
3. We need to verify if the given values satisfy both equations.

STEP 2

1. Substitute the given values into the first equation and check for equality.
2. Substitute the given values into the second equation and check for equality.
3. Conclude whether the given values are a solution to the system.

STEP 3

Substitute x=0 x = 0 and y=13 y = \frac{1}{3} into the first equation:
0+3(13)=1 0 + 3\left(\frac{1}{3}\right) = 1
Simplify:
1=1 1 = 1
The first equation is satisfied.

STEP 4

Substitute x=0 x = 0 and y=13 y = \frac{1}{3} into the second equation:
−3(0)−3(13)=−15 -3(0) - 3\left(\frac{1}{3}\right) = -15
Simplify:
0−1=−15 0 - 1 = -15
−1≠−15 -1 \neq -15
The second equation is not satisfied.

STEP 5

Since the given values do not satisfy the second equation, x=0 x = 0 and y=13 y = \frac{1}{3} are not a solution to the system of equations.

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