Solve the system of two linear equations.

- 30 - 4y = —23
3z - 5y = 5 (1 point)

1 answer

To solve the given system of equations, we first rewrite them clearly:

  1. Equation 1: \(-30 - 4y = -23\)
  2. Equation 2: \(3z - 5y = 5\)

Let's start with the first equation and solve for \(y\):

Step 1: Solve the first equation for \(y\)

\[ -30 - 4y = -23 \]

Add 30 to both sides:

\[ -4y = -23 + 30 \]

Simplifying gives:

\[ -4y = 7 \]

Now, divide by -4:

\[ y = -\frac{7}{4} \]

Step 2: Substitute \(y\) into the second equation

Now we substitute \(y = -\frac{7}{4}\) into the second equation:

\[ 3z - 5y = 5 \]

Substituting for \(y\):

\[ 3z - 5\left(-\frac{7}{4}\right) = 5 \]

This simplifies to:

\[ 3z + \frac{35}{4} = 5 \]

Step 3: Solve for \(z\)

To get rid of the fraction, multiply the whole equation by 4:

\[ 4(3z) + 35 = 20 \]

This simplifies to:

\[ 12z + 35 = 20 \]

Now, subtract 35 from both sides:

\[ 12z = 20 - 35 \]

This simplifies to:

\[ 12z = -15 \]

Now, divide both sides by 12:

\[ z = -\frac{15}{12} = -\frac{5}{4} \]

Summary of Solutions

We've solved the system of equations:

  • \(y = -\frac{7}{4}\)
  • \(z = -\frac{5}{4}\)

Thus, the solution to the system of equations is:

\[ \boxed{\left( y = -\frac{7}{4}, z = -\frac{5}{4} \right)} \]