### Video Transcript

Use the elimination method to solve the given simultaneous equations: negative three π₯ plus four π¦ equals six and negative two π₯ minus three π¦ equals negative 13.

Our first step is to make either the π₯- or the π¦-coefficients the same. In this case, multiplying the top equation by two and the bottom equation by three makes the π₯-coefficient negative six. When we multiply the top equation by two, we end up with negative six π₯ plus eight π¦ equals 12. Multiplying the bottom equation by three gives us negative six π₯ minus nine π¦ equals negative 39.

If we then subtract the two equations to eliminate the π₯ terms, weβre left with 17π¦ equals 51 as eight π¦ minus negative nine π¦ is equal to 17π¦ and 12 minus negative 39 is equal to 51. Dividing both sides of this equation by 17 gives us an answer for π¦ equal to three. We then need to substitute π¦ equals three into one of the equations. In this case, weβre gonna substitute it into equation number one. This gives us negative six π₯ plus eight multiplied by three equals 12.

As eight multiplied by three is 24, we have negative six π₯ plus 24 equals 12. We can add six π₯ to both sides of the equation and then subtract 12 from both sides of the new equation. This leaves us with 12 equals six π₯ or six π₯ is equal to 12. Dividing by six gives us a final answer for π₯ equal to two.

Therefore, the solution to the simultaneous equations negative three π₯ plus four π¦ equals six and negative two π₯ minus three π¦ equals negative 13 are π₯ equals two and π¦ equals three. We could check the solution by substituting the values into equation two. Negative six multiplied by two minus nine multiplied by three is equal to negative 39. Therefore, our solution is correct.