Sometimes, a pair of simultaneous equations may contain one linear and one quadratic. In order to solve this type of equation, we begin with the linear equation.

In the linear equation, the substitution method is used, whereby one of the unknown is made the subject of the formula.

The value got is thereafter re-substituted in the other equation which is quadratic, in two unknowns to obtain a simpler quadratic equation in one unknown.

The quadratic equation in one unknown is then solved using any method, to obtain one or more values.

The value(s) are then substituted again in any of the original two equations to obtain the one or two values for the other unknown.

Having been familiar with the required steps for solving simultaneous, linear, and quadratic equation, let’s solve some examples

Solve the equation

x+y=5—(1)

x^2-2y^2=1—(2)

Make x the subject of the equation in equation 1

x=5-y—(3)

Substitute x = 5y into equation 2

(5-y)^2-y^2=1

Letâ€™s expand the bracket first

5×5-yx5-yx5-yx-y-2y^2=1

25-5y-5y+y^2-2y^2=1

25=10y+y^2-2y^2=1

25=10y-y^2= 1

How? y^2-2y^2=-? y^2 1-2 = -1

Take 25 to the right hand side

-10y-y^2=1-25

-10y-y^2=-24

In order to rearrange the equation, multiply each term by -1

-1x-10y-1x-y^2=-1x-24s

+10y+y^2=+24

y^2+10y-24=1—(4)

Solve by factorization i.ey^2 x-24=-24y^2

Middle term is 10y two factors whose sum gives 10y and product gives -24y^2

Are +12y and-2y

We replace 10y with the two factors ie+12y and-2y in equation 4

(y^2+12y)-(2y-24)=0

y(y^ +12)-2(y+12)=0

The factors are

(y+12)(y-2)=0

y+12=0 and y-2=0

y=-12 or y=2

Substitute the two values of y into equation 3 to obtain the two values of x

x=5-y

When y=-12

x=5-(-12)=5+12=17

When y = 2

x=5-2=3

Therefore, y = -12, x = 17 and y = 2, x = 3

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**Example 2:**

Solve the equation

x=3y=2 and x^2+2y^2=3 Simultaneously

Solution

x=3y=2—(1)

x^2+2y^2=3—(2)

Make the subject of the equation in equation 1

x-3y=2

x^ =2+3y—(3)

Substitute x=2+3y into equation 2

(2+3y)^2+2y^2=3

Let expand the bracket

2×2+2x3y+3yx2+3yx3y+2y^2=3

4+6y+6y+9y^2+2y^2=3

4+12y+11y^2=3

11y^2+12y=-1

11y^2+12y+1=0

Now we have a quadratic equation let solve it by factorization

(11y^2+11y)+(y+1)=0

11(y^ +1)+(y+1)=0

(y^ +1)(11y+1)=0

y+1=0 and 11y+1=0

y=-1 and 11y/11=(-1)/11

y=(-1)/11

Substitute y= -1 and y=(-1)/11 into equation 3

x=2+2y

When y = -1

x=2+3x-1

x=2-3=-1

When y = -1 x= -1

When y=(-1)/11

x=2+3x (-1)/11

x=2+((-3)/11)

x=2+2 (-3)/11

x=11×2/11-3=22/11-3= 19/11

x=1 8/11

x=-1,y=-1 and

y=(-1)/(-1),x=1 8/11

**Example 3:**

Solve the equation

x+2y=2 and x^2+2xy=8

Solution

x+2y= 2—(1)

x^2+2xy=8—(2)

Make x the subject in equation 1

x+2y= 2

x=2-2y—(3)

Substitute equation 3 into equation 2

(2-2y)^2+2(2-2y)y= 8

Let expand the bracket

2×2-2yx2-2yx2-2yx-2y

4-4y-4y+4y^2

4-8y+4y^2

Again, 2(2-2y)y

2(2y-2y^2 )=4y-4y^2

Add them together

4-8y+4y^2+4y-4y^2=8

4-8y+4y+4y^2-4y^2=8

4-4y=8

N/B: 4y^2-4y^2=0

-4y=8-4

-4y/(-4y)=8/(-4)

y=-1

Substitute y = -1 into equation 3

x=2-2y

x=2-2(-1)

x=2-(-1)

x=2+2=4

Therefore, x=4,y=1

**Example 4:**

Solve the equation2x+y=5 and x^2+y^2=25, simultaneously

Solution

2x+y=5—(1)

x^2+y^2=25—(2)

Make y the subject in equation 1

2x^ +y^ =5

y=5-2x—(3)

Substitute equation (3) into equation 2

x^2+(5-2x)^2 = 25

x^2+25-20x+4x^2 = 25

x^2+4x^2+25-20x=25

5x^2-20x=25-25

5x^2-20x=0

Divide through by

5x^2/5-20x/5=0

x^2-4x=0

x^ (x-4)=0

That means x =0 and x-4=0, x = 4

Therefore, x =0 or x= 4

Substitute the values of x into equation 3

y=5-2x

When x =0

y=5-2(0)=5-0

y=5

When x = 4

y=5-2×4

y=5-8=-3

Therefore, the solution is (0,5) and (4, -3)

From what you have learnt in this topic, you can see it with me that simultaneous linear and the quadratic equation is not difficult to understand if you follow the procedure required. I am optimistic that you can tackle any problem involving one-linear one quadratic simultaneous equation.