Last updated: September 29, 2021

Simultaneous quadratics contains two equations, one linear equation and one quadratic equation.

### Linear equation

An equation containing two variables, $$x$$ and $$y$$, form a linear equation if the highest degree of both the variables is $$1$$.

Linear equations are of two types:

• Linear equation in one variable

• Linear equation in two variables

The standard form of linear equation in one variable is $$ax+b=0$$, where $$a$$ and $$b$$ are real numbers. The number $$a$$ is the coefficient, and $$b$$ is the constant of the given equation.

The standard form of linear equation in two variables is $$ax+by+c=0$$ where $$a$$ and $$b$$ are not equal to $$0$$. Here, $$a$$, $$b$$, and $$c$$ are real numbers. The numbers $$a$$ and $$b$$ are the coefficients, and $$c$$ is the constant of the given linear equation.

Any equation whose highest degree is $$2$$ is called a quadratic equation. The standard form of quadratic equation is $$ax^{2} +bx+c=0$$. Here, $$a$$, $$b$$, and $$c$$ are the real numbers, and $$x$$ is an unknown variable. $$a$$ is the coefficient of $$x^{2}$$ and $$b$$ is the coefficient of $$x$$. Here, $$a$$ and $$b$$ represent the coefficients, and $$c$$ represents the constant of the quadratic equation.

## E.5: Derive and solve simultaneous equations involving one linear and one quadratic.

Simultaneous equations can be used to solve various types of problems.

Let us understand the concept of simultaneous quadratics equations through an example:

Suppose $$y=x+3$$ and $$y=x^{2} +3x$$

First, put the value of $$y$$ from the first equation to the second equation.

$$x+3=x^{2}+3x$$

Simplify the above equation.

$$x^{2}+2x-3=0$$

Now, factorise the above equation.

$$\left (x+3 \right) \left (x-1 \right) =0$$

Equate the above equation to zero.

$$x+3=0$$ or $$x-1=0$$

Now, $$x+3=0$$ gives $$x=-3$$, and $$x-1=0$$ gives $$x=1$$.

Therefore, we get either $$x=-3$$ or $$x=1$$.

Substitute the value of $$x$$ in equation $$y=x+3$$.

When we put $$x=-3$$ in the above equation, it gives $$y=0$$

When we put $$x=1$$ in the above equation, it gives $$y=4$$.

Therefore, $$(x,y)=(-3,0)$$ and $$(x,y)=(1,4)$$ are the solutions to the given simultaneous quadratics.

### Worked examples of solving simultaneous quadratics

Example 1: Find the values of $$x$$ and $$y$$ from the system of equations $$y=x-2$$ and $$y=x^{2} +5x+2$$.

Step 1: Substitute the value of $$y$$ from the first equation to the second equation.

$$x-2=x^{2}+5x+2$$

Step 2: Subtract $$x-2$$ from both sides.

$$x^ {2} +4x+4=0$$

Step 3: Solve the above equation by using factorisation.

$$x^ {2} +4x+4=0$$

$$x^{2} +2x+2x+4=0$$

$$x\left (x+2 \right) +2\left (x+2 \right)=0$$

$$\left (x+2 \right) \left (x+2 \right)=0$$

$$\left (x+2 \right)^{2}=0$$

Step 4: Find the value of $$x$$.

$$\left (x+2 \right)^{2}=0$$ gives $$x=-2,-2$$

Step 5: Put the value of $$x in$$y=x-2$$. When we put$$x=-2$$, then$$y=-4$$. Therefore,$$x=-2$$and$$y=-2$$. Example 2: Find the values of$$x$$and$$y$$from equations$$y=-x+5$$and$$y=x^ {2}-7x+14$$. Step 1: Put the value of$$y$$in the second equation.$$ x^{2}-7x+14=-x+5$$Step 2: Subtract$$-x+5$$from both sides.$$ x^{2}-6x+9=0$$Step 3: Solve the above equation by using factorisation.$$x^ {2}-6x+9=0x^ {2}-3x-3x+9=0x\left (x-3 \right)-3\left (x-3 \right)=0\left (x-3 \right) \left (x-3 \right)=0\left (x-3 \right)^{2}=0$$Step 4: Find the value of$$x$$.$$\left (x-3\right)^{2} =0$$gives$$x=3,3$$. Step 5: Substitute the value of$$x\$ in $$y=-x+5$$.

Substitute $$x=3$$ to get $$y=2$$.

Therefore, $$x=3$$ and $$y=2$$.