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Solving a system of equations by substitution of values
simultaneous equations, algebra problem, system of equations
A system of equations requires as many equations as there are variables in the equations.
One way to solve a system is to substitute values successively in each equation to eliminate variables.
Example:
Solve for x and y
Equation (1) 2x+y=16
Equation (2) 3(y+x)=33
There are two variables and we do have two equations so the system should have a solution.
We can find the value of y in terms of x in equation 1 by subtracting 2x from both sides of Equation (1)..
y=16-2x
We may then substitute this value for y into Equation (2). Equation (2) will then look like this:
3[(16-2x) + x]=33
Removing the parentheses by distributing the 3 we get:
48-6x+3x=33
Simplifying the expression Equation (2) now looks like this:
3x=15
This means that x=5.
We may then substitute the value 5 for x in either equation. Substituting 5 for x in Equation 1 we get:
2(5) + y=16.
To solve for y, we subtract 10 from both sides of the equation, leaving us with:
y=6.
We have solved for both x and y and we may then substitute both values into both equations to check our work.
simultaneous equations, algebra problem, system of equations