0 = 5( x + 1 )( 3x + 4 )
zeros: x = -1 and x = -4/3
Answer:
Solving systems of equations with 3 variables is very similar to how we solve systems with two variables. When we had two variables we reduced the system down
to one with only one variable (by substitution or addition). With three variables
we will reduce the system down to one with two variables (usually by addition),
which we can then solve by either addition or substitution.
To reduce from three variables down to two it is very important to keep the work
organized. We will use addition with two equations to eliminate one variable.
This new equation we will call (A). Then we will use a different pair of equations
and use addition to eliminate the same variable. This second new equation we
will call (B). Once we have done this we will have two equations (A) and (B)
with the same two variables that we can solve using either method. This is shown
in the following examples.
Example 1.
3x +2y − z = − 1
− 2x − 2y +3z = 5 We will eliminate y using two different pairs of equations
5x +2y − z = 3
Step-by-step explanation:
The answer is 30
7(8)+4(2)-2(17)=30
It would be 224 , the base is 64 and every triangle is 80 but if you follow the rule for area to a triangle you get 40
So 64+40+40+40+40= 224
