Substitution method : Show
The equations are
Solve for y. Substitute the values
Substitute the values
The solution of the system is
System of Linear Equations entered :[1] 3x - 2y = 14 [2] 5x - y = 32 Graphic Representation of the Equations :-2y + 3x = 14 y + 5x = 32 Solve by Substitution :// Solve equation [2] for the variable y [2] y = 5x - 32 // Plug this in for variable y in equation [1] [1] 3x - 2•(5x-32) = 14 [1] -7x = -50 // Solve equation [1] for the variable x [1] 7x = 50 [1] x = 50/7 // By now we know this much : x = 50/7 y = 5x-32 // Use the x value to solve for y y = 5(50/7)-32 = 26/7
Solution : {x,y} = {50/7,26/7}
System of Linear Equations entered :[1] 3x - 2y = 14 [2] 5x + y = 32 Graphic Representation of the Equations :-2y + 3x = 14 y + 5x = 32 Solve by Substitution :// Solve equation [2] for the variable y [2] y = -5x + 32 // Plug this in for variable y in equation [1] [1] 3x - 2•(-5x+32) = 14 [1] 13x = 78 // Solve equation [1] for the variable x [1] 13x = 78 [1] x = 6 // By now we know this much : x = 6 y = -5x+32 // Use the x value to solve for y y = -5(6)+32 = 2
Solution : {x,y} = {6,2}
How do you find the solution to a system of linear equations?How do I solve systems of linear equations by substitution?. Isolate one of the two variables in one of the equations.. Substitute the expression that is equal to the isolated variable from Step 1 into the other equation. ... . Solve the linear equation for the remaining variable.. What is the solution to the system of equations y =Summary: The solution to the system of equations y = -5x + 3 and y = 1 is x = 0.4 and y = 1.
How many solutions does this linear system have Y 5x 1 15x 3y 3?Summary: If y = 5x - 1 and -15x - 3y = 3, the solution of the linear equations is (0, -1) i.e., one solution.
What are the 3 solutions to systems of equations?There are three methods used to solve systems of equations: graphing, substitution, and elimination. To solve a system by graphing, you simply graph the given equations and find the point(s) where they all intersect.
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