A system can contain as many equations as you can input. Furthermore, it can contain different types of equations. For example, one equation can be linear, and the other can be a nonlinear equation. We will call this system a nonlinear system of equations. In easy words, a system of equations is nonlinear when at least one of its equations is not in the first degree. The condition is that one of the system's equations shouldn't be linear. However, it can have many nonlinear equations but if we talk about the lower limit, the system should have at least one nonlinear equation.

\left\{\begin{matrix} { x }^{ 2 } + { y }^{ 2 } = 25 \\ x + y = 7 \end{matrix}\right

The solution of these systems is usually done by the substitution method. To execute this method, follow the following steps:

1. Work out the value of an unknown in one of the equations, preferably in an equation of the first grade.

y = 7 - x

2. Substitute the value of the unknown in the other equation.

{ x }^{ 2 } + { (7 - x) }^{ 2 } = 25

3.Solve the resulting equation.

{ x }^{ 2 } + 49 - 14x + { x }^{ 2 } = 25

2{ x }^{ 2 } - 14x + 24 = 0

{ x }^{ 2 } - 7x + 12 = 0

x = \frac { -(-7) \pm \sqrt { 49 - 48 } }{ 2 } = \frac { 7 \pm 1 }{ 2 }

x = \frac { 7 + 1 }{ 2 } \qquad x = \frac { 7 - 1 }{ 2 }

x = \frac { 8 }{ 2 } \qquad x = \frac { 6 }{ 2 }

x = 4 \qquad x = 3

 

4. Each of the values obtained are substituted into the other equation, and corresponding values of another unknown are obtained.

{ x }_{ 1 } = 3 \qquad { y }_{ 1 } = 7 - 3 = 4

{ x }_{ 2 } = 4 \qquad { y }_{ 2 } = 7 - 4 = 3

 

Example

\left\{\begin{matrix} x + y = 7 \\ x y = 12 \end{matrix}\right

y = 7 - x

Plugging the value of y in the second equation:

x y = 12

x (7 - x) = 12

7x - { x }^{ 2 } = 12

{ x }^{ 2 } - 7x + 12 =0

x = \frac { -(-7) \pm \sqrt { 49 - 48 } }{ 2 } = \frac { 7 \pm 1 }{ 2 }

x = \frac { 7 + 1 }{ 2 } \qquad x = \frac { 7 - 1 }{ 2 }

x = \frac { 8 }{ 2 } \qquad x = \frac { 6 }{ 2 }

x = 4 \qquad x = 3

 

{ x }_{ 1 } = 4 \qquad { y }_{ 1 } = 7 - 4 = 3

{ x }_{ 2 } = 3 \qquad { y }_{ 2 } = 7 - 3 = 4

 

\left\{\begin{matrix} { x }^{ 2 } + { y }^{ 2 } = 169 \\ x + y = 17 \end{matrix}\right

x = 17 - y

Plugging the above equation in the first equation:

{ (17 - y) }^{ 2 } + { y }^{ 2 } = 169

289 - 34y + { y }^{ 2 } + { y }^{ 2 } = 169

2 { y }^{ 2 } - 34y + 120 = 0

{ y }^{ 2 } - 17y + 60 = 0

y = \frac { -(-17) \pm \sqrt { 289 - 240 } }{ 2 } = \frac { 17 \pm 7 }{ 2 }

y = \frac { 17 + 7 }{ 2 } \qquad y = \frac { 17 - 7 }{ 2 }

y = \frac { 24 }{ 2 } \qquad y = \frac { 10 }{ 2 }

{ y }_{ 1 } = 12 \qquad { y }_{ 2 } = 5

 

{ y }_{ 1 } = 12 \qquad { x }_{ 1 } = 17 - 12 = 5

{ y }_{ 2 } = 5 \qquad { x }_{ 2 } = 17 - 5 = 12

 

 

\left\{\begin{matrix} { y }^{ 2 } - 2y + 1 = x  \\ \sqrt { x } + y = 5 \end{matrix}\right

\sqrt { { y }^{ 2 } - 2y + 1 } + y = 5

{ (\sqrt { { y }^{ 2 } - 2y + 1}) }^{ 2 } = { (5 - y) }^{ 2 }

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

8y = 24

y = 3

 

{ y }^{ 2 } - 2y + 1 = x

{ (3) }^{ 2 } - 2(3) + 1 = x

9 - 6 + 1 = x

x = 4

 

\left\{\begin{matrix} \frac { 1 }{ { x }^{ 2 } } + \frac { 1 }{ { y }^{ 2 } } = 13 \\ \frac { 1 }{ x } - \frac { 1 }{ y } = 1 \end{matrix}\right

\frac { 1 }{ x } = u \qquad \frac { 1 }{ y } = v

\left\{\begin{matrix} { u }^{ 2 } + { v }^{ 2 } = 13 \\ u - v = 1 \end{matrix}\right

u = 1 + v

{ (1 + v) }^{ 2 } + { v }^{ 2 } = 13

1 + 2v + { v }^{ 2 } + { v }^{ 2 } = 13

{ v }^{ 2 } + v - 6 = 0

{ v }^{ 2 } + v(3 - 2) - 6 = 0

{ v }^{ 2 } + 3v - 2v - 6 = 0

v(v + 3) - 2(v + 3) = 0

(v - 2)(v + 3) = 0

v = 2 \qquad v = -3

 

\frac { 1 }{ { x }_{ 1 } } = 3 \qquad { x }_{ 1 } = \frac { 1 }{ 3 }

\frac { 1 }{ { x }_{ 2 } } = -2 \qquad { x }_{ 2 } = - \frac { 1 }{ 2 }

 

\frac { 1 }{ { y }_{ 1 } } = 2 \qquad { y }_{ 1 } = \frac { 1 }{ 2 }

\frac { 1 }{ { y }_{ 2 } } = -3 \qquad { y }_{ 2 } = - \frac { 1 }{ 3 }

The best Maths tutors available
1st lesson free!
Intasar
4.9
4.9 (23 reviews)
Intasar
£42
/h
1st lesson free!
Matthew
5
5 (17 reviews)
Matthew
£25
/h
1st lesson free!
Dr. Kritaphat
4.9
4.9 (6 reviews)
Dr. Kritaphat
£49
/h
1st lesson free!
Paolo
4.9
4.9 (11 reviews)
Paolo
£25
/h
1st lesson free!
Petar
4.9
4.9 (9 reviews)
Petar
£27
/h
1st lesson free!
Rajan
4.9
4.9 (11 reviews)
Rajan
£15
/h
1st lesson free!
Farooq
5
5 (13 reviews)
Farooq
£35
/h
1st lesson free!
Myriam
5
5 (15 reviews)
Myriam
£20
/h
1st lesson free!
Intasar
4.9
4.9 (23 reviews)
Intasar
£42
/h
1st lesson free!
Matthew
5
5 (17 reviews)
Matthew
£25
/h
1st lesson free!
Dr. Kritaphat
4.9
4.9 (6 reviews)
Dr. Kritaphat
£49
/h
1st lesson free!
Paolo
4.9
4.9 (11 reviews)
Paolo
£25
/h
1st lesson free!
Petar
4.9
4.9 (9 reviews)
Petar
£27
/h
1st lesson free!
Rajan
4.9
4.9 (11 reviews)
Rajan
£15
/h
1st lesson free!
Farooq
5
5 (13 reviews)
Farooq
£35
/h
1st lesson free!
Myriam
5
5 (15 reviews)
Myriam
£20
/h
First Lesson Free>

Word Problem

The product of two numbers is 4, and the sum of their squares 17. What are these numbers?

\left\{\begin{matrix} x . y = 4 \\ { x }^{ 2 } + { y }^{ 2 } = 17 \end{matrix}\right

x = \frac { 4 }{ y }

 

{ x }^{ 2 } + { y }^{ 2 } = 17

{ (\frac { 4 }{ y }) }^{ 2 } + { y }^{ 2 } = 17

\frac { 16 }{ { y }^{ 2 } } + { y }^{ 2 } = 17

16 + { y }^{ 4 } = 17{ y }^{ 2 }

{ y }^{ 4 } - 17{ y }^{ 2 } +16 = 0

Let { y }^{ 2 } = t:

{ t }^{ 2 } - 17t +16 = 0

{ t }^{ 2 } - t(16 + 1) +16 = 0

{ t }^{ 2 } - 16t + t +16 = 0

t(t - 16) + 1(t +16) = 0

(t + 1)(t - 16) = 0

t = -1 \qquad t = 16

{ y }^{ 2 } = -1 \qquad { y }^{ 2 } = 16

\sqrt { { y }^{ 2 } } = \sqrt { -1 } \Rightarrow Not \quad Possible \qquad \sqrt { { y }^{ 2 } } = \sqrt { 16 }

y = \pm 4

y = 4 \qquad y = -4

 

x = \frac { 4 }{ y }

For y = 4 \qquad y = -4 :

x = \frac { 4 }{ 4 } \qquad x = \frac { 4 }{ -4 }

x = 1 \qquad x = -1

Need a Maths teacher?

Did you like the article?

1 Star2 Stars3 Stars4 Stars5 Stars 3.00/5 - 2 vote(s)
Loading...

Hamza

Hi! I am Hamza and I am from Pakistan. My hobbies are reading, writing and playing chess. Currently, I am a student enrolled in the Chemical Engineering Bachelor program.