On week 17 of Math 10, we moved onto our last unit, Systems of Linear Relations. For this unit, we are learning how to deal with equations and lines on a graph however this time we are learning how to deal with lines that intersect at a point. For this blog post, which will be the last for Math 10, we only need to know 3 key terms before moving on to our examples. The first word is the Solution which is the exact point of coordinates where two lines intersect. Pretty easy. Second is the Process of Substitution. This is a process where you substitute variables in an equation with the proper coordinate values to find the solution of two lines. The last key term is the Process of Elimination. This is when you add two equations together to find the value of either x or y and then use substitution to find the other coordinate value.
For our first example we are going to find the solution of two lines/equations using the Process of Substitution. Our first example is to find the solution of the equations 6y – 3x + 4 = 0 and 3y = 3x – 5. 
So to begin we are going to isolate one variable from each equation to get the process underway. So between the two equations, It looks like it is easier to isolate the y variable from the second equation (3y = 3x – 5) as all we have to do is divide the whole equation by three. So after we divide 3y = 3x – 5 by 3 and we end up with y = x – 
So what we do is literally input y and to isolate x to end up with the value of the x coordinate. So to do this we distribute substitute the y with x – 
Thirdly what we do is input x into one of the starting equations. Again I used the same equation so we start off with 6y – 3(2) + 4 = 0. We do the same this as we did for finding x which is to isolate the y variable. To do this we sort like terms on opposite sides of the equations and once we do this we end up with 6y = 6 – 4. We then simplify and get 6y = 2. Once again we divide both sides of the equation by 6 and we end up with y = 
Lastly what we do is we verify to check if we did this correctly. So because we used the same equation twice, we might as well do it again. So to begin we start verifying the solution by inputing the x and y values; 6(
Next we are going to find the solution of two equations using the Process of Elimination. This time, we are adding the two equations together to cancel out one term using zero pairs to find our solution. 
So for this example we are going to use the equations x + 3y = 10 and 5x – 3y = 14. Next we are going to see if one term when added together will for a zero pair and in this case we don’t have to do anything because we have one (3y and -3y). After that we add both equations together +$latex \frac{x + 3y = 10}{5x – 3y = 14} to end up with 6x = 24. Lastly we divide both sides of the equation by 6 and end up with x = 4. 
Next, to find y, we do the same as we did with the Process of substitution by substituting x with it’s value which is 4. In this case I chose what looked like the easier equation (x + 3y = 10) and we start off with 4 + 3y = 10. Next we isolate the y variable by sorting like terms on opposite sides of the equation and we get 3y = 10 – 4. After we simplify the equation to get 3y = 6 and we divide both sides by 3 to get y = 2. Lastly we are once again going to verify the solution however this time we are going to use both equations to do this. First I chose to input the solution in the easier equation (x + 3y = 10) to get 4 + 3(2) =10. We then simplify to get 4 + 6 = 10 and we know that both sides of the equation are equivalent. Next we are using the harder equation (5x – 3y = 14) and we get 5(4) – 3(2) = 14. We once again simplify the equation to get 20 – 6 = 14 and we can see that both sides of the equation are equivalent.