In a system of equations, if neither of the equations have an isolated variable (e.g., they are both in standard form), you must start by isolating one of the variables in one of the equations in order to be able to use substitution to solve the system. Choose the variable that would be the easiest to solve for, one that has a coefficient of 1. This way, you won't need to do too many steps in order to isolate the variable. After isolating a variable using inverse operations, plug that value into the other equation and solve. Remember that the solution is a point, so make sure to find both the x and y value of the coordinate point. Check your answer by plugging the x and y values into both equations. If the equations are true, then the solution is correct.
I’m asked to use substitution to solve the system of equations and I’m kind of bummed up because substitution is not so bad if you have one of the variables isolated. In both of these equations, no variable is isolated. Isolated mean like y equals blah, blah, blah, or x equals blah, blah, blah. But when I’m looking for what equation I’m going to have to isolate, or what variable I’m going to isolate and get by itself, I’m going to look at the co-efficient. Coefficients are the numbers dependent on the variables. I want to look for a coefficient of 1 that’s going to make my solving process the most easy and probably reduce fractions if I had any fractions.
Okay so looking here, I can see that that y has a co-efficient of 1. That means I’m going to re-arrange this formula to isolate y, and then I’ll be able to do my substitution. By adding 2x to both sides, I’m not changing this equation, I’m just rewriting it in a form where y is all by itself.
Now I’m going to substitute 2x plus 8 in right there. The whole expression 2x plus 8 is going to get substituted into that second equation. I’m trying to use color in a way that will show you guys what I’m talking about. This raspberry or purplish, reddish color thing is going to be in there for a while. 2 times 2x plus 8. That’s the substitution piece.
Once that’s all done, it’s just solving. I have 1 equation and 1 variable, so just be really careful that you distribute properly meaning that 2 gets multiplied by the 2x, and also by that 8, combine like terms , and then you’re just happily solving along. Subtract 16 from both sides, so 7x equals to -7, and x equals to -1. Again that’s just half of my answer. My answer needs to have an x value and a y value, so I’m not done solving yet. I still have to do some more other problem before I begin checking.
To check, or excuse me to find the y value I’m going to take x equals to -1 and substitute it into either original equation to find my y value. Some people are tempted to plug in their x value into this which should be the equivalent statement, equivalent equation of this first guy, but if I made any kind of error, that’s going to throw off my answer for y.
So instead of plugging into here, I’m going to plug it into either one original equation just to make sure I’m doing everything correctly. So -2 times my x number which was -1 plus y is going to be equal to 8. Go ahead and solve that +2 plus y equals 8, so y equals 6. I think that’s my answer. Before I move on though this problem asked me to check, and it’s always a good idea when you’re doing lots of Algebra like this to check your solution and make sure you didn’t make any mistakes.
The way to check your solution to a system of equations is to plug this x, y pair into each equation separately and make sure I get equalities. Like for the first equation, -2 times -1 plus 6, I hope is equal to 8 and I got that from this first equation. +2 plus 6 equals 8, good that worked.
Let’s try the second equation. 3 times my x number plus 2 times my y number should be equal to 9. Let’s do that out simplifying -3+12=9 good. That means I got the right answer. My x minus y coordinates pair. This was a solution to both original equations, meaning this is where the lines would cross. I didn’t have to graph them, which is great, because I don’t like graphing. I didn’t have to graph them, but I was still able to tell where the lines would intersect.
So one last thing to leave you with, when you see a problem that asks you to use substitution, but no variable is all by itself, look at the coefficients. Find a variable that has a coefficient of 1 and then solve for that guy like we did here.
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