x+27=14

y=3x-14
and
15x+5y=20
y=8-5x

1 answer

To solve the system of equations, let's start with the first pair of equations given:

  1. \( x + 27 = 14 \)
  2. \( y = 3x - 14 \)

Step 1: Solve for \(x\)

From the first equation, we can isolate \(x\):

\[ x + 27 = 14 \implies x = 14 - 27 \implies x = -13 \]

Step 2: Solve for \(y\)

Now we substitute \(x = -13\) into the second equation to find \(y\):

\[ y = 3(-13) - 14 = -39 - 14 = -53 \]

So from the first pair, we have:

\[ x = -13 \quad \text{and} \quad y = -53 \]

Now let's move on to the second pair of equations:

  1. \( 15x + 5y = 20 \)
  2. \( y = 8 - 5x \)

Step 3: Solve the second pair of equations

We can substitute equation 4 into equation 3:

\[ 15x + 5(8 - 5x) = 20 \]

Step 4: Distributing and solving

Now we distribute \(5\):

\[ 15x + 40 - 25x = 20 \]

Combine like terms:

\[ -10x + 40 = 20 \]

Now, isolate \(x\):

\[ -10x = 20 - 40 \implies -10x = -20 \implies x = 2 \]

Step 5: Solve for \(y\)

Now substitute \(x = 2\) back into equation 4 to solve for \(y\):

\[ y = 8 - 5(2) = 8 - 10 = -2 \]

So from the second pair, we have:

\[ x = 2 \quad \text{and} \quad y = -2 \]

Final Results:

From the first pair of equations:

  • \(x = -13\)
  • \(y = -53\)

From the second pair of equations:

  • \(x = 2\)
  • \(y = -2\)