To solve the system of equations using the elimination method, we are given two equations:

1. $-3y + 7x = 15$
2. $-3y + 11x = 21$

Step 1: Subtract the equations

Since both equations have $-3y$ as the $y$-term, we can eliminate the $y$-term by subtracting one equation from the other.

Subtract equation 1 from equation 2: $(-3y + 11x) - (-3y + 7x) = 21 - 15$ Simplifying both sides: $-3y + 11x + 3y - 7x = 6$ This simplifies further to: $4x = 6$

Step 2: Solve for $x$

Now, solve for $x$: $x = \frac{6}{4} = \frac{3}{2}$

Step 3: Substitute $x = \frac{3}{2}$ into one of the original equations

We can substitute $x = \frac{3}{2}$ into either of the original equations. Let’s use the first equation: $-3y + 7x = 15$ Substitute $x = \frac{3}{2}$: $-3y + 7 \times \frac{3}{2} = 15$ Simplifying: $-3y + \frac{21}{2} = 15$ To eliminate the fraction, multiply the entire equation by 2: $2(-3y + \frac{21}{2}) = 2(15)$ Simplifying: $-6y + 21 = 30$ Now, solve for $y$: $-6y = 30 - 21$ $-6y = 9$ $y = \frac{9}{-6} = -\frac{3}{2}$

Final Solution:

The solution to the system of equations is: $x = \frac{3}{2}, \quad y = -\frac{3}{2}$

Would you like further details on any of these steps?

5 Related Questions:

1. Can the substitution method also be used to solve this system?
2. How do you solve systems of equations graphically?
3. What happens if the system of equations has no solutions?
4. How would the elimination method change if there were no common terms?
5. Can elimination be used for systems of three or more equations?

Tip: Always check your solution by substituting the values back into both original equations!