What is the solution to the system of two equations shown? 5x+4y=12 3x+6y=8 you need to find the x and y values, but i have no idea how to.

(x, y) = (2 2/9, 2/9)

Step-by-step explanation:

No doubt you have been exposed to at least a couple of different ways to solve systems of equations like this. The usual methods are elimination and substitution. Either way, you get into annoying fractions with this one.

Elimination

Multiply the second equation by the opposite of the x-coefficient in the first equation, and multiply the first equation by the x-coefficient in the second equation. Add the results.

... 3(5x +4y) -5(3x +6y) = 3(12) -5(8)

... 15x +12y -15x -30y = 36 -40 . . . . . . . . eliminate parentheses

... -18y = -4 . . . . . . collect terms

... y = -4/-18 = 2/9 . . . . . divide by the coefficient of y, and reduce the fraction

Now, put this value into either equation and solve for x.

... 5x +4(2/9) = 12

... 5x = 12 -8/9 = 11 1/9 = 100/9

... x = (1/5)·(100/9) = 20/9 = 2 2/9

___

Comment on Elimination

Here, the solution is messy no matter how you do it, so the method described above works as well as any. For some combinations of coefficients, there are easier ways, but the method described will always work.

Substitution

Solve one of the equations for one of the variables, then use that expression in place of that variable in the other equation.

Solve the first equation for y:

... 4y = 12 -5x . . . . . . subtract the term that doesn't contain y

... y = (12 -5x)/4 = 3 -5/4x . . . . . divide by the coefficient of y

Now, substitute this expression for y into the second equation.

... 3x +6(3 -5/4x) = 8

... 3x +18 -30/4x = 8

... -18/4x = -10 . . . . collect terms, subtract 18

... x = (-10)(-4/18) = 40/18 = 20/9 = 2 2/9

Then putting this value into the expression for y, we can find y to be ...

... y = 3 -5/4(20/9) = 27/9 -25/9 = 2/9

___

Comment on Substituion

As with Elimination, sometimes the set of equations will lend itself to this method. Most often, substitution can be used effectively if you already have an expression for one of the variables (typically, y = ...). If not, sometimes messy fractions are involved with this method, so Elimination can be more appealing.

00

step-by-step explanation: