Simple equations, are equation where you have to find a single unknown.

When an equation contains two unknown quantities it has an infinite number of possible solutions.

When two equations are available connecting the two unknown quantities then a unique solution is possible. Same is true, for three unknowns and three equation connecting the three unknowns, and so on.

Equations which can be solved together to find the unknowns quantities, which are right/ true for all equations involved are called Simultaneous Equations

There are basically two methods to find the solution to simultaneous equations. I.e Substitution and Elimination methods.

We also use graphical methods and determinants of matrices to solve Simultaneously equations. But that we be for my next article on this very topic. In this article, we will be looking at elimination, substitution methods only.

**Example 1: solve the following equations for x and y , (a) by substitution (b) by elimination:**

**(a) By substitution methods**; from the first equation (1): x+2y= -1; make x , subject of formula by subtracting 2y from both sides of the first equations (1)

substitute x, in the second equation of image one I.e , equation (2); 4x-3y=18; by replacing x with this expression (-1-2y), if done correctly you should have;

To remove the bracket, 4 must multiple all values in the bracket. I.e, 4 X -1, to get -4 and 4 X -2y, to get -8y; if done correctly you should have;

to eliminate -4 on the right side equation, add 4 to both sides of the equation

it should be very easy from here, to eliminate -11; divide both sides by -11

substitute y for -2 in equation (1) at the first image. I.e replace y with -2 in this equation: x+2y = -1; if have done it correctly you should have;

to remove -4 on the right side of the equation, add 4 to both sides of the equation

thus, **x= 3**, and **y= -2 **is the solution to the simultaneous equations; x+2y= -1; and 4x-3y=18

**(b) solution by elimination method;**

multiply equation (1) all through by 4 to make the coefficient of x in both equations same, if done correctly you should have;

subtract equation (3) from equation (2)

Note: for the above solution to be possible, 4x – 4x you will get 0; -3y -8y you will get -11y; and 18 -(-4); becomes 18+4; you will get 22.

Next>> substitute y =-2, in equation (1) I.e **-2** should replace y in equation (1): x+2y= -1;

Again, **x= 3**, and **y= -2 **is the solution to the simultaneous equations; x+2y= -1; and 4x-3y=18

**Example 2: Solve the equations on the image below simultaneously using substitution methods**

This equations solution is easiest if a substitution is initially made;

therefore the equations becomes;

thus we have the arrangement below as our final equations to work with;

in equation (4): **a – 4b = -2**; make **a** the subject of formula by adding **4b** to both sides of the equation (4), if done correctly you should have;

now, substitute **-2+4b** to replace **a **in equation (3): **2a+3b = 7** thus, you should have the values below if done correctly;

to remove the -4 on the right side of the equation, add 4 to both sides of the equations and sum them up, if done correctly you should have;

substitute** 1 to replace b** in equation (4): **a-4b = -2, **if done Right your values should look like mine;

to eliminate, -4 on the right side of the equation, add 4 to both sides of the equations and sum them up, if done correctly you should have;

Thus, a = 2 and b = 1, is not the solution to the simultaneous equations yet, we have one more step to go. Remember that **1/x = a** and **1/y = b, therefore, **1/x = 2 and 1/y = 1 ; Cross multiply 1= 2x and 1= y. Thus, x = 1/2 and y =1 is the solution to the simultaneous equations on the image below.

**Example 3: find the solution to the simultaneous equations below**

first eliminate the fractions in both equations;

To eliminate the fractions in equation (1) multiply by the L.C.M of equation (1) ; multiply by 27(x+y)

To eliminate the fractions in equation (2) multiply by the L.C.M of equation (2) ; multiply by 33(2x-y);

the final arrangement of both equations should like what I have here on my image below

since the coefficient of y in both equations are the same, we simply apply **elimination method **by taking the addition of both equations

Note: for the above solution to be possible, 27+33 you will get 60; 4x + 8x you will get 12x; and +4y +(-4y), which is same as 4y-4y, and you will get 0.

Next>> substitute 5 to replace x in equation (3): 27= 4x + 4y

thus, **x=5** and **y= 1 3/4 **is the solution to the simultaneous equations on the image below.

**Practices problem: 1**, the velocity** v **of vehicleis given by the formula **v=u+at . if v = 20 when t = 2 **and** v = 40 when t = 7, **find the values of **u **and** a .** Hence find the velocity of the vehicle** when t = 3.5 .**

**verify your answers with mine**

**[ u = 12, a = 4 , v = 26 ]**

**practices problem 2: **solve the simultaneous equations on the images and verify your answers with mine.

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