Let's solve the system of equations using the substitution method. The given system is:

1. $x - 3y = 10$
2. $x + 5y = -22$

Step 1: Solve one equation for one variable.

We'll start by solving the first equation for $x$:

$x - 3y = 10 \quad \text{(Equation 1)}$

Solve for $x$:

$x = 3y + 10$

Step 2: Substitute the expression for $x$ into the second equation.

Now, substitute $x = 3y + 10$ into the second equation:

$x + 5y = -22 \quad \text{(Equation 2)}$

Substitute $x$ in:

$(3y + 10) + 5y = -22$

Step 3: Simplify and solve for $y$.

Now combine like terms:

$3y + 5y + 10 = -22$

$8y + 10 = -22$

Now, subtract 10 from both sides:

$8y = -32$

Finally, divide by 8:

$y = -4$

Step 4: Substitute $y = -4$ back into the expression for $x$.

Now substitute $y = -4$ into $x = 3y + 10$:

$x = 3(-4) + 10$

$x = -12 + 10$

$x = -2$

Final Answer:

The solution to the system of equations is $x = -2$ and $y = -4$.

Verification:

Let's substitute $x = -2$ and $y = -4$ into the original equations to verify:

1. $x - 3y = -2 - 3(-4) = -2 + 12 = 10$ (True)
2. $x + 5y = -2 + 5(-4) = -2 - 20 = -22$ (True)

Thus, the solution is correct: $x = -2$, $y = -4$.

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Would you like any further details on this method or have any questions?

Related Questions:

1. How would the method change if the system involved three equations?
2. What happens if there are infinite solutions or no solution in a system?
3. How can substitution be used with fractional coefficients?
4. What is the graphical interpretation of the solution to a system of equations?
5. How would you solve a system using the elimination method instead of substitution?

Tip:

To make substitution easier, always choose the equation that allows for easier isolation of a variable, such as one with a coefficient of 1 or -1.