X and Y Table Solver

Introduction :
Solver:

A solver is a generic term indicating a piece of mathematical software, possibly in the form of a stand-alone computer program or as a software library that 'solves' a mathematical problem. A solver takes problem descriptions in some sort of generic form and calculates their solution. (source : Wikipedia)

x and y table solver:

The solver that used to solve the system of linear equations called as x and y table solver.

In this article we are going to see some sample and practice problems on x and y table solver.


Solved problems on x and y table solver:

Problem 1:

Solve the following system of equation by using x and y table solver.

5x + 8y = -6

3x - 4y =14

Solution:

Given,   5x + 8y = -6

3x - 4y  =14

The given equations are system of linear equation.

We can solve these equations by using substitution method or elimination method.

Here we are going to solve these equations by using both methods.

5x+8y = -6 --------- > 1

3x-4y=14  ---------- > 2

Multiply the equation 2 by 2,

2(3x-4y) = 2(14)

6x - 8y = 28 ----------- > 3

Add the equation 1 and 3,

5x+8y = -6

(+)6x - 8y = 28

11x -0 = 22

11x = 22

Divided by 11 on both sides,

x =22/11

x = 2

Substitute x =2 in the equation 1,

5x+8y = -6

5(2)+8y = -6

10+8y = -6

Subtract 10 on both sides,

10 + 8y - 10 = -6 -10

8y = -16

Divide by 8 on both sides,

y = -16/8

y = -2

The solution of the system of linear equation is x =2 and y = -2

Answer: x = 2 , y = -2

Problem 2:

Solve the following system of equation by using x and y table solver.

x + y =10

8x- y = 8

Solution:

Given,   x + y =10

8x- y = 8

The given equations are system of linear equation.

We can solve these equations by using substitution method or elimination method.

Here we are going to solve these equations by using both methods.

x + y =10 --------- > 1

8x- y = 8  ---------- > 2

Add the equation 1 and equation 2,

x + y = 10

(+)   8x - y = 8

9x = 18

Divide by 9 on both sides,

9x/9 =18/9

x = 2

Substitute x = 2 in 1,

x + y =10

2 + y =10

Subtract 2 on both sides,

2 + y -2 = 10 -2

y = 8

The solution of the system of given linear equation is x =2 and y = 8

Answer: x = 2 , y =8

Practice problems on x and y table solver:


Problems:

1.Solve the following system of equation by using x and y table solver.

x + y = 150

x - y = - 50


2. Solve the following system of equation by using x and y table solver.

x + 3y = 36

3x + y =  44

Answer key:

1. x =50 and y =100

2. x =12 and y = 8

Study Online Linear

Introduction :

Online Linear equation is incorporated form an algebraic equation, for example y = 5x + 8 or 5x + 6y – 2z = 4. The degree of the variable of the linear equation is 1. If the Online linear equation has two variable, that  online  linear equation will give  the straight line on the graph. online is one of the best tool to study a lot of things.

Procedure to study online linear equation:


Step 1:  If we want to solve the equation for x, we  have to move the x to on side of the equal sign, which  is similar to,    10x+10+50=300

Step 2:Do the 'reverse operations.' This means we would want to add -50.  So we will get 10x+10=250

Step 3: Now we have to add -10 on both sides of the equation. So we will get 10x=240

Step 4: Now we have to divide both sided by 10. That is 10x/10=240/10. Therefore, x=24

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Examples to study online linear problems:


Ex:1

x - 4 = 10

Add 4 to both sides of the equation:

x = 14

The answer is x = 14

Ex:2

2x - 1 = 10

Add 1 to both sides of the equation:

2x = 11

Divide both sides by 2:

X = 5.5

The answer is x = 5.5

Ex::

x+y=9

-x+2y=0

Substitution method for linear equation:

x+y=9   ---------------------- equation 1
-x+2y=0---------------------- equation 2

If we add the equations 1 and 2, we will get

3y=9

3y/3 = 9/3 ( both sides are divided by 3 )

y = 3

Substitute y = 3 in the equation 1, so we will get

x +  3 = 9

x+3-3=9-3 ( -3 is added on both sides)

x=6

Elimination method for linear equation:

x+y=9   ---------------------- equation 1
-x+2y=0---------------------- equation 2

Take the equation 1

x+y=9

x+y-y=9-y ( -y is added on the both sides )

x=9-y

Substitute x=9-y  in the equation 2, we will get

-(9-y ) +y=0

-9+y+2y=0

-9+3y=0

-9+9+3y=0+9

3y=9

3y/3=9/-3 ( both sides are divided by 3)

Y=3

Substitute y=3 in the equation 1

X+3=9

X+3-3=9-3 ( Add -3 on the both sides)

X=6

Solving Online Simultaneous Equations

Introduction to online simultaneous Equation:

A set of linear equations having a common solution set is called a system of simultaneous linear equation. In solving simultaneous equation section we are going to learn how to solve the linear equations in three variables To find the values of three unknowns, we need to be given three linear equations on three unknown variables.In general online is defined as tutoring on internet. Thus the solving of  simultaneous equation which is taught on internet is known as solving online simultaneous equation.

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Procedure for solving three given linear equations with three variables x, y, z


Three equations are given.
Take any two, say the first two equations.
Eliminate one variable say z.
Similarly eliminate z from the second and the third (or first and the third equation).
We get two linear equations in x, y.
Solve them in the usual way learnt in early cases.
Substitute the values of x and y in any one of the three equations to get the value of z.
Thus the values of x, y and z are obtained.

Please express your views of this topic linear equations in one variable word problems by commenting on blog.

Example Problems on online simultaneous equation:


Example 1: Solve : x+ y = 3, y + z = –5, z + x = 2.

Solution: Let the equations be identified as

x + y = 3------------------- (1)

y + z = –5 ------------------(2)

z + x = 2 --------------------(3)

Adding all the three equations, we get

2x + 2y + 2z = 3 + (–5) + 2 or 2(x + y + z) = 0 or x + y + z = 0 (4)

Substituting y + z = –5 in equation (4) we get x + (–5) = 0 ∴x = 5.

Substituting z + x = 2 in equation (4) we get y + 2 = 0 ∴ y = –2.

Substituting x = 5 in equation (3) we get z + 5 = 2 or z = 2–5 = –3

The solution is x = 5, y = –2, z = –3.

Example 2: solve: x +3y = 4, x + 6y = 2.

Solution: Let the equation can be written as

X + 3y = 4 ------------(1)

X + 6y = 2 ------------(2)

Changing the sign      __-__-____-___

-3y   =  2

Y = -(2/3).

Sub y in (1) we get

X + 3{-(2/3)} = 4

X – 2 = 4

X = 6.

Setting up Equations for Given Situations

Learning setting up equations for given situations:

In different situations we may need to find sum of x and y such that sum of x and y are equal to 4 or difference between any two variables equal to 9. Instead of describing the situations in words, we may wish to set them up in mathematical equations using mathematical operations like +, -, x, or /.The words sum, total, altogether,added,... denotes the mathematical symbol +. The words difference, subtract, taken away,... denotes the mathematical symbol -. The words product, multiplied,... denotes the mathematical symbol x.


Problems on Setting up Situations:

In this article let us learn setting up equations for given following simple situations:

1. Sum of x and y is 67. Equation is x + y = 67

2. Take away y from x to form 8. Equation is x - y = 8

3. Product of a and b is 8. Equation is ab = 8

4. Quotient of x and y is 5. Equation is x/y = 5

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More Problems on Setting up Situations:

Let cost of a pen is $x and cost of a pencil is $y.Frame an equation such that 37 pens and 53 pencils together cost $890

Equation is 37x + 53y = 890

In a two digit number, the unit's digit is twice the ten's digit.

Equation is x = 2y where x is the digit in the unit's place and y is the digit in the ten's place.

A two digit number decreased by 54, the digits are reversed.

Equation is 10x+y - 54 = 10y+x

The sum of a two digit number and the number formed by interchanging its digits is 110.

Equation is 10y + x + 10x + y = 110

Area of a rectangle gets reduced by 9 square units if its length is reduced by 5 units and breadth is increased by 3 units.

Equation is xy - 9 = (x - 5)(y + 3)

Solving Linear Equations Substitution Method

Introduction :

The linear equations for the substitution methods are one of the methods in the systems of equations in the algebra section. The substitution methods are nothing but the variable can be found by using the equations and the value of one variable can be determined. Then, other variable can be found by substituting the variable value in another equation. Now we see the solving of linear equations by substitution method.

About Solving of Linear Equations by Substitution Method:

Now we are going to see the steps for solving the linear equations by substitution methods are as follows,

The equation should be choose and place one variable in one side and the other variables and values on the other side.

Then, in the second step these equations should be substituted in one equation and the value can be determined.

The other value of the variable can be determined by substituting the value in another equation.

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Problems for Solving the Equations by Substitution Method:

Example 1:

Solve the given two equations by substitution method:

x + y = 2

y + 10 = 2x

Solution:

Choose one of the equations for solving the linear systems as follows,

x + y = 2

x = 2 – y

Now put the value of x in the equation y + 10 = 2x

y + 10 = 2x

y + 10 = 2(2 – y)

y + 10 = 4 – 2y

y + 2y = 4 - 10

3y = -6

y = -2.

Put the value of y in the equation x = 2 - y

x = 2 - y

x = 2 - (-2)

x = 2 + 2

x = 4

The values are x = 4 and y = -2.

Example 2:

The given two equations for solving linear system are:

x + y = 4

y + 8 = x

Solution:

Choose one of the equations for solving the linear systems as follows,

x + y = 4

x = 4 - y

Now put the value of x in the equation y + 8 = x

y + 8 = x

y + 8 = 4 - y

y + y = 4 - 8

2y = - 4

y = -2.

Put the value of y in the equation x = 4 - y

x = 4 - y

x = 4 - (-2)

x = 4 + 2

x = 6

The values are x = 6 and y = -2.