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toc = Simultaneous Equations =

Recall that to ** solve simultaneous equations ** means to find the coordinates of the point where their graphs intersect.

At this point, both equations are true at the same time.

Some simultaneous equations can be solved by plotting both graphs carefully on graph paper (See: Solving simultaneous equations using graphs).

OR

We can use technology (graphing program or CAS calculator) to solve simultaneous equations.

OR

We can use algebra.

We will learn two algebraic methods:
 * Substitution Method (use when one of the equations is in the form ** y = mx + c ** )
 * ** Elimination Method ** (use when both the equations are in the form ** ax + by = c ** )

Elimination Method
This method involves 5 steps:
 * 1) Manipulate one or both of the equations so that the coefficients of y (or x) are the same in both equations
 * 2) Add or Subtract the two equations so that the y term (or the x term) is eliminated.
 * 3) This gives an equation purely in terms of x, which we can solve for x. (or purely in terms of y, which we can solve for y)
 * 4) Substitute the value of x into one of the original equations and solve for y (or the other way)
 * 5) State the solution as a set of coordinates ** (x, y) **

** Example 1 **
Solve the following simultaneous equations using the elimination method:
 * 5x + 2y = 18
 * 3x – 2y = –2

__**Solution:**__



__**Checking Solution:**__

{We can check our solution is correct by substituting ** (2, 4) ** into equation ** [2] **: the equation we didn't use to find y}

math \\ . \qquad 3x-2y=-2 \qquad \text{Substitute } \big(2,\;4\big) \\ \\ . \qquad 3 \times 2 - 2 \times 4 = -2 \\ \\ . \qquad 6 - 8 = -2 math

LHS = RHS so solution is correct. **(2, 4)** ü

** Example 2 **
Solve the following simultaneous equations using the elimination method:
 * 2x + 2y = 4
 * 6x + 3y = 15

__**Solution:**__



__**Checking Solution:**__

{Substitute ** (3, –1) ** into equation ** [2] **: the equation we didn't use to find y}

math \\ . \qquad 6x+3y=15 \qquad \text{Substitute } \big(3,\;-1\big) \\ \\ . \qquad 6 \times 3 + 3 \times ^-1 = 15 \\ \\ . \qquad 18 + ^-3 = 15 math

LHS = RHS so solution is correct. **(3, –1)** ü

** Example 3 **
Solve the following simultaneous equations using the elimination method:
 * 5x + 2y = 15
 * 2x – 3y = 6

__**Solution:**__ __**Checking Solution:**__

{Substitute ** (3, 0) ** into equation ** [2] **: the equation we didn't use to find y}

math \\ . \qquad 2x-3y=6 \qquad \text{Substitute } \big(3,\;0\big) \\ \\ . \qquad 2 \times 3 - 3 \times 0 = 6 \\ \\ . \qquad 6 - 0 = 6 math

LHS = RHS so solution is correct. **(3, 0)** ü

Note: The most common error when solving simultaneous equations, is to stop after finding the x and forget to find the y-value as well. Remember that you are finding the coordinates of a point so you need both x and y values. State your solution as coordinates each time.

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