All Operations in the Order of Operations: More than Four Numbers

Due to the fact that the exercise only involves addition and subtraction operations, we will solve it from left to right:

$12+6=18$

$18-4=14$

$14+18=32$

$32-12=20$

Due to the fact that the exercise only involves addition and subtraction operations, we will solve it from left to right:

$14-5=9$

$9-9=0$

$0+7=7$

$7+2=9$

Due to the fact that the exercise only involves addition and subtraction operations, we will solve it from left to right:

$21-3=18$

$18-6=12$

$12+9=21$

$21-5=16$

Due to the fact that the exercise only involves addition and subtraction operations, we will solve it from left to right:

$25-6=19$

$19-9=10$

$10+7=17$

$17-3=14$

Due to the fact that the exercise only involves addition and subtraction operations, we will solve it from left to right:

$26-6=20$

$20+9=29$

$29+7=36$

$36-12=24$

Due to the fact that the exercise only involves addition and subtraction operations, we will solve it from left to right:

$32-4=28$

$28-19=9$

$9+3=12$

$12-7=5$

Insofar as the exercise only involves addition and subtraction operations, we will solve it from left to right:

$30+6=36$

$36-5=31$

$31+7=38$

$38-17=21$

According to the rules of the order of arithmetic operations, we must first solve the multiplication exercises.

We place them inside of parentheses in order to avoid confusion during the solution:

$2+(3\times6)-(3\times7)+1=$

We then solve the multiplication exercises:

$2+18-21+1=$

Lastly we solve the rest of the exercise from left to right:

$2+18=20$

$20-21=-1$

$-1+1=0$

According to the order of operations, we will first solve the multiplication exercises.

We will put them in parentheses so that we don't get confused later in the solution:

$25-(3\times4)+(4\times2)=$

Let's solve the multiplication exercises:

$3\times4=12$

$4\times2=8$

We get:

$25-12+8=$

Let's solve the exercise from left to right:

$25-12=13$

$13+8=21$

According to the rules of the order of arithmetic operations, we must first enclose both the multiplication and division exercises inside of parentheses:

$1+(2\times3)-(7:4)=$

We then solve the exercises within the parentheses:

$2\times3=6$

$7:4=\frac{7}{4}$

We obtain the following:

$1+6-\frac{7}{4}=$

We continue by solving the exercise from left to right:

$1+6=7$

$7-\frac{7}{4}=$

Lastly we break down the numerator of the fraction with a sum exercise as seen below:

$7-(\frac{4+3}{4})$

$7-(\frac{4}{4}+\frac{3}{4})$

$7-(1+\frac{3}{4})$

$7-1\frac{3}{4}=5\frac{1}{4}$