how is solving a system of equations algebraically? How is it different?
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Answer:
I am assuming that you are referring to a system of linear equations in two variables. If that is, indeed, the case, then you can either solve by substitution (solve one of the equations for one variable in terms of the other), then substitute that expression into the other equation, or you can solve by elimination (that is, adding or subtracting the two to eliminate one of the variables).
As an example, suppose the two equations are 5x + 2y = 30 and 2x – y = 3.
To solve by substitution, solve the second equation for y (y=2x – 3), then substitute the expression 2x – 3 for y in the other equation, 5x + 2(2x – 3)=30. Solve for x, then substitute into either equation to find the value of y:
5x + 4x – 6 = 30; 9x = 36; x=4. y = 2(4) – 3 so y = 8 – 3; y = 5 Solution {(4,5)}
To solve by elimination: 5x + 2y = 30
multiply other equation by 2 so that the y variables in both are opposites:
4x – 2y = 6
Add the equations: 9x = 36, so x= 4. Substitute into either equation to find the corresponding value of y : 5(4) + 2y = 30; 20 + 2y = 30; 2y = 10; y=5
Solution is {(4,5)}.
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