Algebra

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Simultaneous Equations: Elimination 2

Simultaneous Equations: Elimination 2

Building on the first elimination lesson, learn how to solve simultaneous equations when their coefficients are different.

Elimination is a method of solving simultaneous equations. Sometimes, you will have to multiply one or both equations before they can be added together to eliminate one of the variables.

Have a look at $-3x + 2y = 5$ and $6x + 5y = 8$

Neither the $x$ or $y$ coefficients here add to make 0. We need to multiply the equations first to be able to eliminate one of the variables.

Let's have a go at solving $-3x + 2y = 5$ and $6x + 5y = 8$ simultaneously

Label the equations 1 and 2

$(1)-3x + 2y = 5$ and $(2)\space6x + 5y = 8$

Multiply the equations to make a variable add to 0

The goal is to multiply one or both of the equations so that when we add them together, either $x$ or $y$ adds together to give 0, and therefore allows us to find the other.

Multiply equation (1) by $2$ to equalise $x$

$2\times(-3x+2y=5)$ becomes $-6x+4y=10$

Add $-6x+4y=10$ and $6x + 5y = 8$

$4y +5y=10+8$

$9y = 18$

$9y = 18$ , so what is y?

Put $y=2$ into equation 1

$-3x+2(2)=5$

Solve $-3x+2(2)=5$ to find $x$ as a fraction

Nice! You've solved it

$x=-\dfrac{1}{3}$ and $y=2$

Solve $2x + y = 5$ and $3x - 2y = 4$

Try solving $3x + y = 10$ and $2x - 3y = 14$

Sometimes you will be given two equations where both may have to be multiplied by different numbers before a variable can be eliminated.

Look at $3x + 2y = 4$ and $2x + 3y = 6$ . In order to solve, we need to eliminate $y$

Label equations 1 and 2

$(1)\space3x + 2y = 4$ and $(2)\space2x + 3y = 6$

Multiply equation (1) by 3

$3\times(3x + 2y = 4)\rightarrow9x+6y=12$

Multiply equation (2) by -2

$-2\times(2x + 3y = 6)\rightarrow-4x-6y=-12$

Add the equations together

$(9x+6y=12)+(-4x-6y=-12)\rightarrow5x=0$

$5x=0$ , so what is $x$ ?

Put $x=0$ into equation (1)

$3(0)+2y=4$

$3(0)+2y=4$ , so what is y?

$y=2$

Nice! We have solved the simultaneous equations. $x=0$ and $y=2$ .

Have a go at solving these: $4x + 3y =17$ and $3x - 4y=-6$

Now try $3x + 2y = 17$ and $2x + 5y = 26$