This week in Math 10, we learnt to solve systems using substitution. Solving by substitution is an algebraic way of solving a system. Substitution is just inserting one equation into another equation and isolating a variable.

For this example I’ll use the equations; x+4y=-3 and 3x-7y=29.

As we can see, we don’t know what x or y is. With some rearranging, we can at least figure out the equation that a variable could be. In the first equation, x+4y=-3, we can move the 4y to the other side to get x by itself. The rearranged equation would now be; x=-3-4y. Now that we know what x is, we can insert it into the second equation. It would now look like; 3(-3-4y)-7y=29.

The first thing we do is distributive property.

-9-12y-7y=29. We now put the like terms together.

-9-19y=29. Next thing is isolation the variable. For that, we move the -9 to the other side which would make it a +9. -19y=29+9.

-19y=38. to fully isolate the variable, we divide everything by -19. Our equation would be y=-38/19 which we would then need to reduce if possible. The final answer is y=-2.

Now that we have y, we need to find x. We can use the easiest equation that we have and replace y with -2.

We will replace the y in the x+4y=-3 equation.

x+4(-2)=-3

x-8=-3. Now we move the -8 to the other side to get x by itself.

x=-3+8 –> x=5.

Now that we have x &y, we have to verify the numbers in both equations.

x+4y=-3 –> (5)+4(-2)=-3 –> 5-8=-3 –> -3=-3

3x-7y=29 –> 3(5)-7(-2)=29 –> 15+14=29 –> 29=29

The numbers worked! Our ordered pair looks like; (5,-2)