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About Josh
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When I work through problems, I emphasize principles and key ideas which I believe are worth noting. I will try to answer questions in the following areas, but not at the advanced level. Algebra. Sequences & Series. Trigonometry. Functions & Graphs. Coordinate Geometry. Quadratic Polynomials. Exponentials & Logarithms. Basic Calculus. Probability, Permutations and Combinations. Mathematical Induction. Complex numbers. Physics problems.

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Experience:
I have worked as a teaching assistant in college. My hope is that more people will share knowledge without boundary, give help without seeking recognition or monetary rewards.

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See a selection of past questions in my maths repository under "Question Archive"

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Bachelor degree in Engineering Science.

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You are here: Experts > Science > Math for Kids > Basic Math > simultaneous linear equations

Basic Math - simultaneous linear equations

Expert: Josh - 11/4/2009

Question
Hi Josh,

y = 2x+3
y = 5x-3            how do i solve this to get x=2 and y=7

and y = 3x -1                     to get x=1 and y=2
    2x+4y = 10

ive tried various ways, elimination,substitution, yet i still come up with different answers.

Answer
Hi Richard,

Given two equations with two unknowns, we can use elimination and back-substitution to find the values of x and y.

When solving simultaneous equations, I like to think of each variable ("x" and "y") as nothing more than a place holder. Algebra arithmetic then works pretty much in the same way as with numbers. For example, if we call y=2x+3 [equation 1], y=5x-3 [equation 2], then we can carry out "elimination" and "back-substitution" in two steps.

Step 1: Elimination of y
Method: compute [2]-[1] (i.e., equation 2 minus equation 1)

As a ground rule, whatever operation we perform on one side of an equation, we do the same to the other side as well. On the left hand side (LHS), y-y gives 0. Eliminating the variable "y" from the equation allows us to get closing to knowing x. On the right hand side, we have (5x-3)-(2x+3). This is same as 5x-3 -2x-3. Collecting the "x" terms together, we have 5x-2x-3-3. (Note: conceptually, 5x-2x is simplified in the same way as 5-2. Taking two lots of "x" from five lots of "x" gives us three lots of "x"). In summary, the RHS of the equation becomes 3x-6. The end result (LHS=RHS) reads 0=3x-6.

Now, our goal is to find x. Once x is known, y can be found. Adding 6 to both sides of the equation "0=3x-6" gives 6=3x. Dividing both sides by 3, we get x=2. Remember: to avoid making mistakes, we must be consistent in what we do, always apply the same operation to both sides of the equation. (i.e., we add/subtract/multiply the same amount to both sides of an equation)

Step 2: Back substitution
Method: this can be done in a number of ways. As long as we plug-in the value of "x" into one of the original equations [2] or [1], we will find the value of "y" which satisfies both equations.

e.g., put x=2 into equation [1], we get y=2*2+3, i.e., y=7.

Q2. For this question, we have y=3x-1 [1] and 2x+4y=10 [2]. We can once again use the "elimination" "back-substitution" technique to solve this. For example, by computing 4*[1]-[2] to eliminate "y" first up. This time, without all the commentary, I'll just show you the working.

Step 1: ELIMINATION
4*[1]-[2] means
4y - (2x+4y) = 4*(3x-1) -10   [note: 4y-4y=0 on the LHS, after simplifying]
-2x = 12x-4 -10 [then, add 2x to both sides of the equation]
0 = 14x -14 [then, add 14 to both sides]
14 = 14x
x = 1.

Step 2: BACK SUBSTITUTION
put x=1 into [1], y=3*1-1 gives y = 2.

Perhaps an easier way to solve this is by direct substitution. i.e., replacing y in equation [2] with 3x-1 (using the relationship stated in [1]). This leads to 2x+4(3x-1)=10. After some algebra (expansion, simplification etc.), we get the same answer.

Cheers!

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