This is not a paper
Answers
x+y=5 [1]
x+y=5 [1]xy=6 [2]
x+y=5 [1]xy=6 [2]To solve for x and y , you should rearrange one of the equations for one of the variables x or y . Choose equation [1] to rearrange for y , because it is easy to do so.
x+y=5 [1]xy=6 [2]To solve for x and y , you should rearrange one of the equations for one of the variables x or y . Choose equation [1] to rearrange for y , because it is easy to do so.y=5−x [3]
x+y=5 [1]xy=6 [2]To solve for x and y , you should rearrange one of the equations for one of the variables x or y . Choose equation [1] to rearrange for y , because it is easy to do so.y=5−x [3]Now, substitute this expression into equation [2]:
x+y=5 [1]xy=6 [2]To solve for x and y , you should rearrange one of the equations for one of the variables x or y . Choose equation [1] to rearrange for y , because it is easy to do so.y=5−x [3]Now, substitute this expression into equation [2]:x(5−x)=6
x+y=5 [1]xy=6 [2]To solve for x and y , you should rearrange one of the equations for one of the variables x or y . Choose equation [1] to rearrange for y , because it is easy to do so.y=5−x [3]Now, substitute this expression into equation [2]:x(5−x)=6 Now use this equation to solve for x . Firstly expand the parentheses:
x+y=5 [1]xy=6 [2]To solve for x and y , you should rearrange one of the equations for one of the variables x or y . Choose equation [1] to rearrange for y , because it is easy to do so.y=5−x [3]Now, substitute this expression into equation [2]:x(5−x)=6 Now use this equation to solve for x . Firstly expand the parentheses:5x−x2=6
x+y=5 [1]xy=6 [2]To solve for x and y , you should rearrange one of the equations for one of the variables x or y . Choose equation [1] to rearrange for y , because it is easy to do so.y=5−x [3]Now, substitute this expression into equation [2]:x(5−x)=6 Now use this equation to solve for x . Firstly expand the parentheses:5x−x2=6 Rearrange the terms so that they are all on one side of the equation:
x+y=5 [1]xy=6 [2]To solve for x and y , you should rearrange one of the equations for one of the variables x or y . Choose equation [1] to rearrange for y , because it is easy to do so.y=5−x [3]Now, substitute this expression into equation [2]:x(5−x)=6 Now use this equation to solve for x . Firstly expand the parentheses:5x−x2=6 Rearrange the terms so that they are all on one side of the equation:x2−5x+6=0
x+y=5 [1]xy=6 [2]To solve for x and y , you should rearrange one of the equations for one of the variables x or y . Choose equation [1] to rearrange for y , because it is easy to do so.y=5−x [3]Now, substitute this expression into equation [2]:x(5−x)=6 Now use this equation to solve for x . Firstly expand the parentheses:5x−x2=6 Rearrange the terms so that they are all on one side of the equation:x2−5x+6=0 What you have here is a quadratic equation. There are many ways to solve a quadratic equation but I will choose the easiest method for this problem.
x+y=5 [1]xy=6 [2]To solve for x and y , you should rearrange one of the equations for one of the variables x or y . Choose equation [1] to rearrange for y , because it is easy to do so.y=5−x [3]Now, substitute this expression into equation [2]:x(5−x)=6 Now use this equation to solve for x . Firstly expand the parentheses:5x−x2=6 Rearrange the terms so that they are all on one side of the equation:x2−5x+6=0 What you have here is a quadratic equation. There are many ways to solve a quadratic equation but I will choose the easiest method for this problem.Factorise the quadratic expression on the left-hand side of the equation by using the product sum method (or by decomposition):
sum method (or by decomposition):(x−2)(x−3)=0
sum method (or by decomposition):(x−2)(x−3)=0 The null factor law gives x=2 or x=3 as the solution.
sum method (or by decomposition):(x−2)(x−3)=0 The null factor law gives x=2 or x=3 as the solution.Now, substitute x=2 or x=3 into equation [3] to get the value of y:
sum method (or by decomposition):(x−2)(x−3)=0 The null factor law gives x=2 or x=3 as the solution.Now, substitute x=2 or x=3 into equation [3] to get the value of y:when x=2 , then y=5−2=3
sum method (or by decomposition):(x−2)(x−3)=0 The null factor law gives x=2 or x=3 as the solution.Now, substitute x=2 or x=3 into equation [3] to get the value of y:when x=2 , then y=5−2=3 when x=3 , then y=5−3=2
sum method (or by decomposition):(x−2)(x−3)=0 The null factor law gives x=2 or x=3 as the solution.Now, substitute x=2 or x=3 into equation [3] to get the value of y:when x=2 , then y=5−2=3 when x=3 , then y=5−3=2 Therefore, there are two solutions to these simultaneous equations, which are: (2,3) and (3,2)