System of Equations
There are two methods of solving systems of equations:
Substitution is where we substitute one equation into the other by isolating a certain variable, or a group of terms.
Elimination is where we subtract the two equations. Before doing this, we may have to multiply one equation by a certain number to make sure one variable cancels out.
Solving the Question
We're given the following equations:
Because they are organized in the same manner (i.e. x [operation] y [equals] number), it is easier for us to use elimination.
First, multiply the first equation by 2:
![x + 3y = 7\\2(x + 3y) = 2(7)\\2x + 6y = 14](https://img.qammunity.org/2023/formulas/mathematics/high-school/z5e9eb3lmkoepkvkpkh4sqtsbh0gimwz3n.png)
Now, subtract the second equation from the one we just created:
![\hspace{10}2x + 6y = 14\\-2x + 4y = 11\\\rule{67}{0.3}\\2y=3](https://img.qammunity.org/2023/formulas/mathematics/high-school/w0wtojmgsit0ki1pjqqwgxd878jsz6bqzw.png)
Solve for y:
![y=(3)/(2)](https://img.qammunity.org/2023/formulas/mathematics/high-school/cef5rar9jex4lturpiokdvmgow9t02f3pu.png)
To solve for x, we can use substitution in the first equation:
![x + 3y = 7\\\\x + 3((3)/(2)) = 7\\\\x + (9)/(2) = 7\\\\x = 7- (9)/(2)\\\\x = 7- 4.5\\\\x = 2.5\\\\x=(5)/(2)](https://img.qammunity.org/2023/formulas/mathematics/high-school/o8xc93mh9fh4wk39jjg3qwu09p1wy60aix.png)
Answer
![x=(5)/(2)](https://img.qammunity.org/2023/formulas/mathematics/high-school/1yegkzvdad42h7kqsd48ouo0746zi8mkyk.png)
![y=(3)/(2)](https://img.qammunity.org/2023/formulas/mathematics/high-school/cef5rar9jex4lturpiokdvmgow9t02f3pu.png)