Algebra Tutorials!

 Wednesday 18th of January

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 Depdendent Variable

 Number of equations to solve: 23456789
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 Dependent Variable

 Number of inequalities to solve: 23456789
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# Algebra - Two Variables

## Variable Substitution

You can solve a system of equations with a method called substitution. Hereâ€™s how to use it.

You have two equations, so â€œsolveâ€ one equation by writing A in terms of B. That is, use the rules of algebra to rearrange the equation to get â€œA =â€ on one side. Then substitute this expression into the other equation wherever variable A appears. This results in one equation and one unknown.

Example:

y = 4x

x + y = 90

What are the values of x and y?

Solution:

 The first equation y = 4x tells you that â€œ4xâ€ is another name for y. Substitute 4x for y: x + y = 90 x + 4x = 90 5x = 90 Divide both sides by 5: 5x/5 = 90/5 x = 18 To find y, substitute 18 for x in either of the original equations: y = 4x y = 4(18) y = 72 Substitute the values of x and y into the other equation: x + y = 90 ? 18 + 72 = 90 ? Yes!

Example:

Solve the following system of two equations:

Q + N = 33.

25Q + 5N = 505.

 Look at the equations. What constant should we use? -5(Q + N) = -5(33) Letâ€™s multiply the second equation by -5: â€“5Q â€“5N = -165 Add both equations together: 25Q + 5N  =  505 -5Q â€“ 5N = -165 20Q =  340 Solve for Q: 20Q/20 = 340/20 Q = 17 Substitute Q = 17 to solve for N: 17 + N = 33 N = 33 â€“ 17 = 16 Check the result with both: 25(17) + 5(16) = 505? Yes! 17 + 16 = 33? Yes!