To solve the system of equations given by
- \( 7x + 10y = 10 \)
- \( -3x - 2y = 14 \)
we can use either the substitution method or the elimination method. Here, I'll use the elimination method.
Let's first rewrite the second equation in a more convenient form:
\[ 3x + 2y = -14 \]
Now we have the two equations:
- \( 7x + 10y = 10 \)
- \( 3x + 2y = -14 \)
Next, we can eliminate one of the variables. To eliminate \( y \), we can make the coefficients of \( y \) in both equations the same. We can multiply the second equation by 5 to get the coefficient of \( y \) to match that in the first equation:
Multiply the second equation by 5:
\[ 5(3x + 2y) = 5(-14) \implies 15x + 10y = -70 \]
Now our equations are:
- \( 7x + 10y = 10 \)
- \( 15x + 10y = -70 \)
Next, we'll subtract the first equation from the second to eliminate \( y \):
\[ (15x + 10y) - (7x + 10y) = -70 - 10 \]
This simplifies to:
\[ 15x - 7x = -80 \implies 8x = -80 \]
Now we solve for \( x \):
\[ x = \frac{-80}{8} = -10 \]
Now that we have \( x \), we can substitute it back into one of the original equations to find \( y \). We can use the first equation:
\[ 7(-10) + 10y = 10 \]
This simplifies to:
\[ -70 + 10y = 10 \]
Now solve for \( y \):
\[ 10y = 10 + 70 \implies 10y = 80 \implies y = \frac{80}{10} = 8 \]
Thus, the solution to the system of equations is:
\[ (x, y) = (-10, 8) \]
The final answer is:
\[ \boxed{(-10, 8)} \]