We talked at the beginning of the unit
about solving systems of equations by graphing, but rarely does that give us a concrete solution...which is why we have algebraic methods! Hooray for algebra! After these few lessons, you should be able to
The first method to discuss is solve by substitution. When we have a system of equations, there is NO way to get x= and number or y=a number. So, we have to do some kind of manipulation to get an equation with ALL x's or ALL y's. The way we do this is by first isolating a variable within one of the equations (either equation, get x or y by itself, it doesn't matter). We then can plug that in for the isolated variable in the OTHER equation, giving us a chance to solve. Once we have the first variable solved, we can plug it back into one of our original equations and solve for the other unknown. Sounds confusing? Look at the examples below.
about solving systems of equations by graphing, but rarely does that give us a concrete solution...which is why we have algebraic methods! Hooray for algebra! After these few lessons, you should be able to
- Solve systems by substitution
- Solve systems by elimination (add and subtract)
- Solve systems by elimination (multiplication)
The first method to discuss is solve by substitution. When we have a system of equations, there is NO way to get x= and number or y=a number. So, we have to do some kind of manipulation to get an equation with ALL x's or ALL y's. The way we do this is by first isolating a variable within one of the equations (either equation, get x or y by itself, it doesn't matter). We then can plug that in for the isolated variable in the OTHER equation, giving us a chance to solve. Once we have the first variable solved, we can plug it back into one of our original equations and solve for the other unknown. Sounds confusing? Look at the examples below.
The main thing to remember is that you need ONE variable in an equation to solve. The way we get that is by substituting in for the second unwanted variable.
We spent two days on the method of elimination. The first day was with simpler problems where you already have a matching term, and the second day was for the cases where there IS no matching term. Let's look at the first (simpler) case:
We spent two days on the method of elimination. The first day was with simpler problems where you already have a matching term, and the second day was for the cases where there IS no matching term. Let's look at the first (simpler) case:
Notice in both cases, we have a matching term (the left example was 1x and -1x, the right example was 1y and -1y). Thus, if we were to combine the equations, that variable would cancel out, leaving us with just ONE to solve. Once you find the one variable, you can follow the same steps as substitution: go back to an original equation, plug in the known variable, and solve for the second unknown.
In both examples, we had one positive and one negative matching term. What happens if they are both positive or both negative? If we were to add the equations, then nothing would cancel. So instead, we SUBTRACT the equations. Let's look at the above right example. Though it is simpler to cancel out the y's, there ARE two positive x's. If we instead wanted to cancel the x's, we would subtract the second equation of x-y=14. However, we're not just subtracting x. We are subtracting the entire equation. Thus, you need to distribute in a negative to EVERY term, resulting in a new equation of -x+y=-14. After that negative is distributed, we can combine the equations like normal and continue solving.
Elimination can get a bit trickier though...what if there are not matching terms at all? In this case, we have to create matching terms through multiplication.
In both examples, we had one positive and one negative matching term. What happens if they are both positive or both negative? If we were to add the equations, then nothing would cancel. So instead, we SUBTRACT the equations. Let's look at the above right example. Though it is simpler to cancel out the y's, there ARE two positive x's. If we instead wanted to cancel the x's, we would subtract the second equation of x-y=14. However, we're not just subtracting x. We are subtracting the entire equation. Thus, you need to distribute in a negative to EVERY term, resulting in a new equation of -x+y=-14. After that negative is distributed, we can combine the equations like normal and continue solving.
Elimination can get a bit trickier though...what if there are not matching terms at all? In this case, we have to create matching terms through multiplication.
Notice that to get a matching term, we multiplied the ENTIRE equation by -3. Why not just 3 you ask? Well, if we multiply by NEGATIVE 3, then we can add the equations and the two cancel right away. If we had multiplied by a positive 3, we'd have to subtract and distribute a negative anyways. Might as well do it while you're multiplying by the factor of 3! Sometimes you have to multiply BOTH equations by something...
I'm attaching quite a few documents. If you would like to fill in the notes, talk to a peer or come see my copy.
-Mrs. Mooney
-Mrs. Mooney
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