Equations that have letters are just called equations, but there are types 1, 2, 3, 4 and ratio type equations. Ratio type equations are the ones where it is a fraction on both sides of the equals sign. Equation (Type 2): x+1-5=12 Ratio Type Equation: 21/7=42-x/3
If you already know that x = -3 and y = 5 what linear equations are you wanting to solve?
Um there are an infinite number of equations but some simple ones are: X + 1 = 6 X + 2 = 7 123553X = 617765
It works out that the points of intersection between the equations of 2x+5 = 5 and x^2 -y^2 = 3 are at: (14/3, -13/3) and (2, 1)
On a graph, parallel equations will run forever and never touch, whereas, perpendicular equations will make a right angle with each other.In an equation, parallel lines will have the same slope (the mx part of the equation).Ex:y = (1/3)x + 2y = (1/3)x + 4Perpendicular equations have negative reciprocal slopes of each other.Ex:y = (3/5)x + 6y = -(5/3)x + 4
Equations that have letters are just called equations, but there are types 1, 2, 3, 4 and ratio type equations. Ratio type equations are the ones where it is a fraction on both sides of the equals sign. Equation (Type 2): x+1-5=12 Ratio Type Equation: 21/7=42-x/3
If you already know that x = -3 and y = 5 what linear equations are you wanting to solve?
That depends. What are the equations?
6-3=3 and 5-2=3 and 4-1=3 and 3-0=3 and 2-(-1)=1
Um there are an infinite number of equations but some simple ones are: X + 1 = 6 X + 2 = 7 123553X = 617765
The answer will depend on statement 3 5 - whatever that may be!
2x + 3 = 7 and x = 2 3y - 5 = 7 and y = 4
It works out that the points of intersection between the equations of 2x+5 = 5 and x^2 -y^2 = 3 are at: (14/3, -13/3) and (2, 1)
On a graph, parallel equations will run forever and never touch, whereas, perpendicular equations will make a right angle with each other.In an equation, parallel lines will have the same slope (the mx part of the equation).Ex:y = (1/3)x + 2y = (1/3)x + 4Perpendicular equations have negative reciprocal slopes of each other.Ex:y = (3/5)x + 6y = -(5/3)x + 4
No.
Do you mean: 4x+7y = 47 and 5x-4y = -5 Then the solutions to the simultaneous equations are: x = 3 and y = 5
true