How can I use approximations to find solutions to systems of equations?

Step-by-step explanation:

Solve a linear system by multiplying one equation.

Solve a linear system of equations by multiplying both equations.

Compare methods for solving linear systems.

Solve real-world problems using linear systems by any method.

Introduction

We have now learned three techniques for solving systems of equations.

Graphing, Substitution and Elimination (through addition and subtraction).

Each one of these methods has both strengths and weaknesses.

Graphing is a good technique for seeing what the equations are doing, and when one service is less expensive than another, but graphing alone to find a solution can be imprecise and may not be good enough when an exact numerical solution is needed.

Substitution is a good technique when one of the coefficients in an equation is +1 or -1, but can lead to more complicated formulas when there are no unity coefficients.

Addition or Subtraction is ideal when the coefficients of either x or y match in both equations, but so far we have not been able to use it when coefficients do not match.

In this lesson, we will again look at the method of elimination that we learned in Lesson 7.3. However, the equations we will be working with will be more complicated and one can not simply add or subtract to eliminate one variable. Instead, we will first have to multiply equations to ensure that the coefficients of one of the variables are matched in the two equations.

Quick Review: Multiplying 

step-by-step explanation:

5×5×5+2(8)÷2(2)+3(3)

15+16÷4+9

31÷13

do 31÷13 and u will get ur answer

x=-7

step-by-step explanation:

x+3=-10

-10+3=-7

x=-7