A few days ago
V for V

Math problem and systems of equations ten easy points?

How would you solve this problem without a calculator.

1=A(x-1) + B

and also this systems of equations without calculators

A+C=0

B-2C=4

-A+B+C=0

Top 3 Answers
A few days ago
Anonymous

Favorite Answer

1= a(x-1) – b

1-b= a(X-1)

(1-B)/a= x-1

x= ((1-b)/a)+1

a+c=0; so a=(-c)

b-2c=4; so b=4+2c

-a+b+c=0; so -(-c)+(4+2c)+c=0

c+4+2c+c=0

4c+4=0

4c=-4

c=-1

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A few days ago
?
You need to solve for each letter then put the answers into the equation to solve

I will not delinquent you by providing the answers…LOL>

The bottom equation solve for C first in both the top two equations by subtituting 0 for the other Letters…Good Luck..

Solve Systems of Equations – Tutorial

This is a tutorial on solving 2 by 2 systems of linear equations. Detailed solutions and explanations are provided.

Example 1: Solve the system of linear equations.

-2x + 3y = 8

3x – y = -5

Solution to example 1

* multiply all terms in the second equation by 3

-2x + 3y = 8

9x – 3y = -15

* add the two equations

7x = -7

* Note: y has been eliminated, hence the name: method of elimination

* solve the above equation for x

x = -1

* substitute x by -1 in the first equation

-2(-1) + 3y = 8

* solve the above equation for y

2 + 3y = 8

3y = 6

y = 2

* write the solution to the system as an ordered pair

(-1,2)

* check the solution obtained

first equation: Left Side: -2(-1) + 3(2)= 2 + 6 = 8

Right Side: 8

second equation: Left Side: 3(-1)-(2)=-3-2=-5

Right Side: -5

* conclusion: The given system of equations is consistent and has the ordered pair, shown below, as a solution.

(-1,2)

Matched Exercise 1: Solve the system of linear equations.

-x + 3y = 11

4x – y = -11

Example 2: Solve the system of linear equations.

4x – y = 8

-8x + 2y = -5

Solution to example 2

* multiply all terms in the first equation by 2

8x – 2y = 16

-8x + 2y = -5

* add the two equations

0x + 0y = 11 or 0 = 11

* Conclusion: Because there are no values of x and y for which 0x + 0y = 11, the given system of equations has no solutions. This system is inconsistent.

Matched Exercise 2: Solve the system of linear equations.

2x + y = 8

-6x – 3y = 10

Example 3: Solve the system of linear equations.

2x – 3y = 8

-4x + 6y = -16

Solution to example 3

* multiply all terms in the first equation by 2

4x – 6y = 16

-4x + 6y = -16

* add the two equations

0x +0y = 0 or 0 = 0

* Conclusion: The system has an infinite number of solutions. The solution set consists of all ordered pairs satisfying the equation 2x – 3y = 8. This system is consistent.

Matched Exercise 3: Solve the system of linear equations.

x – 2y = 3

-3x + 6y = -9

References and links related to systems of equations.

* Systems of Equations Calculator and Solver.

* Systems of Linear Equations – Graphical Approach

* Solve Equations, Systems of Equations and Inequalities.

Home Page – Online Calculators – Trigonometry – Antennas – Graphing – Precalculus Tutorials – Calculus Tutorials

Geometry Tutorials – Precalculus Applets – Applied Math – Precalculus Questions and Problems –

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Updated: 23 January 2006 (A Dendane)

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4 years ago
Anonymous
Steps to doing those issues: a million. Multiply an entire equation by ability of a undeniable fee to make one varible in one equation equivalent to an analogous variable in yet another equation. Subtract the two equations to depart merely one variable and a relentless. sparkling up for the 2d variable, insert it into the equation and sparkling up for the different variable.
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