How To Find X Value In System Of Equations

Systems of Linear Equations

Often it is necessary to look at several functions of the aforementioned independent variable. Consider the prior example where x, the number of items produced and sold, was the independent variable in three functions, the toll office, the revenue function, and the profit role.

In full general there may be:

north equations

v variables

Solving systems of equations

At that place are four methods for solving systems of linear equations:

a. graphical solution

b. algebraic solution

c. emptying method

d. substitution method

Graphical solution

Instance ane

given are the ii following linear equations:

f(x) = y = 1 + .5x

f(x) = y = 11 - 2x

Graph the first equation by finding two data points. By setting first x and then y equal to zero information technology is possible to detect the y intercept on the vertical axis and the ten intercept on the horizontal axis.

If ten = 0, so f(0) = 1 + .5(0) = 1

If y = 0, so f(x) = 0 = 1 + .5x

-.5x = 1

x = -2

The resulting data points are (0,1) and (-ii,0)

Graph the second equation by finding 2 information points. Past setting starting time x and and then y equal to zero it is possible to find the y intercept on the vertical centrality and the x intercept on the horizontal axis.

If x = 0, then f(0) = 11 - two(0) = 11

If y = 0, then f(x) = 0 = 11 - 2x

2x = 11

ten = five.5

The resulting data points are (0,11) and (v.5,0)

At the indicate of intersection of the two equations 10 and y have the same values. From the graph these values can be read as x = 4 and y = three.

Example 2

given are the two following linear equations:

f(x) = y = 15 - 5x

f(x) = y = 25 - 5x

Graph the beginning equation by finding 2 information points. Past setting first x and and so y equal to nothing it is possible to find the y intercept on the vertical axis and the x intercept on the horizontal axis.

If x = 0, then f(0) = 15 - 5(0) = fifteen

If y = 0, and then f(x) = 0 = 15 - 5x

5x = xv

x = 3

The resulting data points are (0,15) and (3,0)

Graph the 2nd equation by finding ii data points. By setting first 10 and then y equal to cypher it is possible to find the y intercept on the vertical axis and the x intercept on the horizontal axis.

If x = 0, then f(0) = 25 - 5(0) = 25

If y = 0, so f(x) = 0 = 25 - 5x

5x = 25

ten = 5

The resulting data points are (0,25) and (5,0)

From the graph it can be seen that these lines do not intersect. They are parallel. They accept the same slope. In that location is no unique solution.

Example 3

given are the two post-obit linear equations:

21x - 7y = 14

-15x + 5y = -x

Rewrite the equations by putting them into slope intercept course.

The first equation becomes

7y = -14 + 21x

y = -ii + 3x

The second equation becomes

5y = -10 + 15x

y = -two + 3x

Graph either equation past finding 2 information points. Past setting get-go x and so y equal to zip information technology is possible to find the y intercept on the vertical axis and the ten intercept on the horizontal axis.

If 10 = 0, then f(0) = -2 +3(0) = -2

If y = 0, and so f(ten) = 0 = -2 + 3x

3x = 2

x = 2/iii

The resulting data points are (0,-2) and (ii/three,0)

From the graph it can exist seen that these equations are equivalent. There are an infinite number of solutions.

Algebraic solution

This method will be illustrated using supply and demand assay. This type of analysis is derived from the work of the peachy English language economist Alfred Marshall.

Q = quantity and P = price

P (s)= the supply role and P (d) = the demand office

When graphing toll is placed on the vertical axis. Thus price is the dependent variable. It might be more logical to call back of quantity equally the dependent variable and this was the arroyo used past the groovy French economist, Leon Walras. However past convention economists proceed to graph using Marshall'southward analysis which is referred to as the Marshallian cross.

The objective is to find an equilibrium price and quantity, i.e. a solution where toll and quantity volition have the same values in both the supply function and the toll function.

Q_Eastward = the equilibrium quantity P_E = the equilibrium price

For equilibrium
supply = need
or P (due south) = P (d)

Given the post-obit functions

P (s) = 3Q + 10 and P (d) = -1/2Q + fourscore

Prepare the equations equal to each other and solve for Q.

P (s) = 3Q + 10 = -1/2Q + 80 = P (d)

iii.5Q = seventy

Q = 20 The equilibrium quantity is 20.

Substitute this value for Q in either equation and solve for P.

P (s) = 3(20) + 10

P (due south) = 70

P (d) = -ane/two(20) + fourscore

P (d) = 70 The equilibrium toll is 70.

Elimination method

This method involves removing variables from the equations. Variables are removed successively until simply a unmarried concluding variable is left, i.e. until at that place is one equation with one unknown. This equation is so solved for the 1 unknown. The solution is then used in finding the second to last variable. The process is repeated by adding back variables equally their solutions are plant.

Example i

2x + 3y = 5

-5x - 2y = 4

Process: eliminate y. The coefficients of y are not the same in the two equations but if they were it would possible to add together the two equations and the y terms would cancel out. However information technology is possible through multiplication of each equation to force the y terms to have the aforementioned coefficients in each equation.

Step i: Multiply the kickoff equation by 2 and multiply the 2nd equation by 3. This gives

4x + 6y = 10

-15x - 6y = 12

Step 2: Add the 2 equations. This gives

-11x = 22

ten = -2

Stride 3: Solve for y in either of the original equations

2(-2) + 3y = 5

3y = 9

y = 3              or

-5(-2) - 2y = 4

x - 2y =     iv

2y =     half dozen

y = 3

Alternate Procedure: eliminate x. The coefficients of 10 are not the same in the ii equations but if they were information technology would possible to add the two equations and the y terms would cancel out. However information technology is possible through multiplication of each equation to force the x terms to have the same coefficients in each equation.

Stride ane: Multiply the first equation by five and multiply the second equation past 2. This gives

10x + 15y = 25

-10x - 4y = 8

Step ii: Add together the ii equations. This gives

11y = 33

y = 3

Step 3: Solve for x in either of the original equations

2x + 3(3) = 5

2x = -4

ten = -two             or

-5x     - two(3) = iv

           - 5x =   10

10 =     -ii

Case 2

2x₁ + 5x₂ + 7x_three = 2

4x₁ - 4x_two - 3x₃ = 7

3x₁ - 3x₂ - 2x₃ = v

In this example there are three variables: x₁, x_two, and ten_three. One possible procedure is to eliminate beginning x_i,to eliminate next 10_two, so to solve for x₃. The value obtained for ten₃ is used to solve for x_iiand finally the values obtained for 10₃ and 10₂ are used to solve for x₁.

Procedure Part A Outset eliminate x_ane.

Step 1 Multiply the first equation by 2 and subtract the 2d equation from the first equation. This gives

4x₁ + 10x₂ + 14x₃ = 4 offset equation

4x₁ - 4x_two - 3x₃ = 7 second equation

14x₂ + 17x_iii = -iii second equation subtracted from the start

Step 2 Multiply the commencement equation by 3, multiply the third equation by ii, and subtract the third equation from the first equation. This gives

6x₁ + 15x₂ + 21x_iii = 6 get-go equation

6x₁ - 6x_two - 4x₃ = ten third equation

21x₂ + 25x_iii = -4 third equation subtracted from the commencement

Process Part B Second eliminate x_two. From Function A at that place are two equations left. From these two equations eliminate ten₂.

14x₂ + 17x₃ = -3 outset equation

21x₂ + 25x₃ = -4 second equation

Footstep i Multiply the first equation by 21, multiply the 2d equation by 14. and subtract the 2nd equation from the commencement equation. This gives

294x_ii + 357x_iii = -63 first equation

294x_two + 350x_three = -56 2nd equation

7x₃ = -7 second equation subtracted from beginning

x₃ = -ane

Part C Solve for x_ii by inserting the value obtained for 10₃ in either equation from Part B.

14x₂ + 17(-1) = -3

1 4x_two = 14

ten₂ = 1 or

21x₂ + 25(-1) = -iv

21x₂ = 21

x_two = one

Part D Solve for x₁ past inserting the values obtained x₂ andx_three in any of the three original equations.

2x_one + 5x₂ + 7x₃ = 2 offset original equation

2x_ane + 5(1) + seven(-ane) = 2

2x_i = 4

x₁ = 2 or

4x_ane - 4x_two - 3x₃ = seven second original equation

4x₁ - four(ane) - 3(-1) = seven

4x₁ = viii

ten₁ = 2 or

3x_ane - 3x_two - 2x_iii = 5 tertiary original equation

3x₁ - 3(one) -2(-1) = 5

3x₁ = 6

x₁ = 2

Substitution method

This involves expressing one variable in terms of another until at that place is a single equation in one unknown. This equation is then solved for that one unknown. The result is then used to solve for the variable which was expressed in terms of the variable whose solution has merely been found.

Case

12x - 7y = 106 first equation

8x + y = 82 2d equation

Solve the 2d equation for y and then substitute the value obtained for y in the offset equation.

y = 82 - 8x second equation solved for y

12x - 7(82 – 8x) = 106 first equation rewritten in terms of x

12x - 574 +56x = 106

68x = 680

x = ten

Substitute the value obtained for ten in either of the original equatioins.

12x - 7y = 106 first equation

12(ten) - 7y = 106

7y = xiv

y = two

8(10) + y = 82 second equation

y = two

[Index]