I would use graphing because I could first make an x/y chart for y=4x+5 x=1, y=9/ x=0, y=5 /x=-1, y=1. Then, I could get y on one side by itself y=1 2/11 -6/11x make an x/y chart x=1, y=7/11/ x=0, y=1 2/11/ x=-1, y=1 8/11. I could then set 4x+5=13/11-6/11x. Solve for x. Substitute the answer for x in each of the original equations and solve for y in each to see if there is an intersecting point.
Cindy Thank you for sharing your work.
You have made life very hard for yourself here. You appear to have been taught three methods for solving systems of simultaneous equations. The question asks you
Without solving, what method would you choose to solve the system: graphing, substitution, or elimination? Explain your reasoning.
y= 4x + 5
6x + 11y = 13
Your answer above mixes up two different methods and does not really answer the question posed.
Let's talk about the graphical method first. You seem to understand that one. You draw the graph of y = 4x + 5 by building an x/y table. Because this is a linear equation, you need to calculate just two arbitrary points, but it is very good practice to calculate three. All three should lie on the same straight line when you join them with a straightedge on graph paper. If they do not, you made a mistake in your x/y table. So far so good. You now do the same thing with the equation 6x + 11y = 13. Build an x/y table for three points, make sure they lie in a straight line, draw the line on the same graph as the line for the other equation. The point where the two lines intersect is the answer. The question does not ask you to do that. It asks you to think about it to determine whether you think that is the easiest method and, if so, to explain why you think it is easier than the other two methods.
You then go off to another method, which is perfectly valid, but not quite one of the methods suggested.
In the substitution method, you use one equation to express one unknown in terms of the other or others. You then rewrite each of the remaining equations to get rid of that unknown. You keep getting rid of unknowns until you have just one equation with one unknown. You solve that. And then you go back and solve for the other unknowns. And at the end, you check everything.
\(\displaystyle 6x + 11y = 13 \implies 6x = 13 - 11y \implies x = \dfrac{13 - 11y}{6}.\)
\(\displaystyle So\ y = 4x + 5 \implies y = 4\left(\dfrac{13 - 11y}{6}\right) + 5 \implies 6y = 4(13 - 11y) + 5 * 6 = \implies 6y = 52 - 44y + 30 \implies 6y + 44y= 52 + 30 \implies 50y = 82 \implies\)
\(\displaystyle y = \dfrac{82}{50} = \dfrac{41}{25} \implies \dfrac{41}{25} = 4x + 5 \implies 41 = 100x + 125 \implies 100x = 41 - 125 = - 84 \implies x = - \dfrac{84}{100} = - \dfrac{21}{25}.\)
Let's check. \(\displaystyle 4\left(- \dfrac{21}{25}\right) + 5 = \dfrac{-84}{25} + \dfrac{125}{25} = \dfrac{41}{25}.\) That checks.
\(\displaystyle 6\left(- \dfrac{21}{25}\right) + 11\left(\dfrac{41}{25}\right) = \dfrac{-126}{25} + \dfrac{451}{25} = \dfrac{325}{25} = 13.\) That checks.
Now here is the thing. This method works for just about any system that can be solved, but it involves a lot of algebra, where you can make mistakes. You have to learn it because it is so general and powerful, but I for one use it only when I have to because I make lots of mistakes in my algebra.
That leaves elimination. This is easy to use if you have integer coefficients and just two linear equations in two unknowns. You multiply one of the equations so that one variable has a coefficient of the same magnitude but opposite signs in the two equations. You add those equations. Now you have one equation in one unknown. You solve that. Now that you know one unknown you use either equation to solve for the other unknown. Then you check.
\(\displaystyle y = 4x + 5.\)
\(\displaystyle 11y = 13 - 6x.\) Multiply the equation above by minus 11.
\(\displaystyle - 11y = - 44x - 55.\) Add this equation and the one above.
\(\displaystyle 0 = 13 - 6x - 44x - 55 \implies 6x + 44x = 13 - 55 \implies 50x = - 42 \implies x = - \dfrac{42}{50} = -\dfrac{21}{25}.\)
\(\displaystyle So\ y = 4\left(- \dfrac{21}{25}\right) + 5 = \dfrac{-84}{25} + \dfrac{125}{25} = \dfrac{41}{25}.\)
We should check now, but we did that before and know these answers are correct. So can you now explain one of the methods and why you prefer it?
