# Simultaneous quadratics

**What are simultaneous quadratics?**

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.

**Quadratic 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.**

$$x^ {2}-6x+9=0$$

$$x^ {2}-3x-3x+9=0$$

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

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

$$\left (x-3 \right)^{2}=0$$

**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 $$y=2$$.