Math Examples of Variables and Expressions

Here we can clearly see that we have an inequality 5 k – 5 > 5 for which we have to find out the solutions where it satisfies the condition and ‘k’ is a variable term. So we put the values of ‘k’ in the left hand side terms and check the condition. In starting we put the value of k = 1 then
5 * 1 – 5 = 5 – 5 = 0,
0 is less than 5 which does not fulfill the condition. So we again put the value k = 2 so,
5 * 2 – 5 = 10 – 5 = 5,
Here we can clearly see that k = 2 we get the value 5 which is equals to 5 but not greater than 5. So condition does not fulfill. We again put the value k = 3 so then in that case
5 * 3 – 5 = 15 – 5 = 10,
Condition fulfill the requirement because at k = 3 then we get value 10 which is greater than 5.

Here we can clearly see that we have an equation 5n – 3 = 17. Where ‘n’ is a variable in which we put the values of integers. So now we put the value n = 1 n the equation and check whether it satisfy the condition or not.
For n = 1 5 * 1 – 3 = 5 – 3 = 2,
The value 2 does not satisfy the condition. So now we again put the value of n which is 2
For n = 2 5 * 2 – 3 = 10 – 3 = 7,
In that case also it does not satisfy the condition. This time we take the value n = 3. So
For n = 3 5 * 3 – 3 = 15 – 3 =12.
12 also not fulfill the condition. At last we take the value n = 4,
For n = 4 5 * 4 – 3 = 20 – 3 = 17
Here we can clearly see that the value 17 fulfill the right hand side condition. So we can say that for the value of z = 4 the given equation satisfy the condition.

As we all know from the property of Open Sentence, for the values of ‘z’ it may be true or false. Here we can clearly see from the inequality 3z – 2 < 5 that z is a variable where we have to put some integers values and find out that for which values of ‘z' it satisfy the condition.
Now we put the value z = 1 in the left hand side term 3 * 1 – 2 = 3 – 2 = 1,
So for the value z= 1 it satisfy the condition which is true. Now we again put the value in ‘z’, if we put the value z = 2 then in that case,
3 * 2 - 2 = 4,
4 is less than 5 which is true statement so it satisfies the condition. Now we put the value z = 3 then
3 * 3 – 2 = 9 – 4 = 5,
In that case it does not fulfill the condition because 5 are equals to 5 but not less than 5. So it is not a true statement. Now it very clear that for all value of ‘z’ which are greater than 1 it does not fulfill the condition and for all values of ‘z’ which are less than 1 it satisfy the condition.

As we all know from the property of open sentence that it should be true or false depend on variable value. Here we can easily see from the inequality 2 + 4x < 6 that x is a variable where we have to put some Integer’s value in it. To check for what value the inequality 2 + 4x < 6 satisfy the condition. So from the inequality
2 + 4x < 6
Now we put the value of x = 1 in the left hand side then 2 + 4 * 1 = 6.
But it is not greater less than 6 so it is not satisfy the condition. And for all values of x which are greater than 1 it also not satisfy the condition. Now if we put value less than 1 means (-) negative integers like -1 then
2 + (-1) * 4 = 2 – 4 = - 2,
In that case we can say that – 2 is less than 6 and for all values ‘x’ which are less than 0 then equation satisfy the condition.

As we all know from the property of open sentence, it may be true or false depends on variable value. Here we can easily see that in the equation 3y – 9 = 21, ‘y’ is a variable where the equation depends. So we put the value of ‘y’ in the equation,
For y = 1,
3 * 1 – 9 = -6,
So for y = 1 condition not fulfill. Now we put the value of y = 2,
For y = 2
 3 * 2 – 9 = - 3,
Again condition does not full fills. So we again put the value of y = 3,
For y = 3,
3 * 3 – 9 = 0,
After trying lots of positive integers value we put the value y = 10
For y = 10,
3 * 10 – 9 = 21
Here we can easily see that equation fulfill the requirement at y = 10. So the equation is true at y = 10 in the open sentence.

Here we can clearly see that expression contains bracket term which is (7x – 3) and the value of ‘x’ is 2. So first we solve that term by putting the value of ‘x’ so
(7 * 2 – 3) = 14 – 3 = 11,
After that we put the value of ‘x’ in 2x term so
2x = 2 * 2 = 4,
At last after solving both ‘x’ terms we put the values in the expression
4 + 11 = 15,
So the final value of the expression 2x + (7x – 3) is 15.

So to solve that expression first we solve the bracket term where we have two values 5 and 2. And between them there is ‘–‘ sign that’s why we subtract 2 from 5 and the resulting value will again subtract with term 2y, after subtracting 2 from 5 we get 3. Now we will put the value of ‘y’ in 2y.
2y = 2 * 5 = 10,
 At last we subtract 3 from 10 so the resulting value,
10 – 3 = 7.

Here we can clearly see that expression contains bracket so first we solve the bracket of the expression where we have to put the value of y = 9. So the bracket term after putting the value of ‘y’
(9 – 6) = 4,
After that we add that value from left most value 31 of an expression with 4 so
31 + 4 = 35.
 Now we let’s take another example where we have an expression 4k + 3k and k = 4. Next we have to find out the value of the expression. To calculate the value first we put the value of k = 4 in term 4k after putting a value we get
4 * 4 =16,
 Now we put the value of k = 4 in a term 3k after putting a value we get,
3 * 4 = 12,
 And at last we add both values 16 and 12 so the resulting value is
16 + 12 =28.

To solve that first we solve the bracket term which is (12 - 4). In which we subtract 11 from 3 so the resulting value we get is 12 – 4 = 8. In Second step we put the value of p=3 in the 6k, terms of the equation after putting the value of k = 6 * 3 = 18. After getting the value of term 6p we add that with 8 which is solution of the expression. So the final result of expression 6p + (12 – 4) is:
18 + 8 = 26.

Here we can easily see that first we have two digit number 211 which we have to add to next term (31 – 11). So to add second term (31 - 11), first we have to solve (31 - 11). After solving that
(31 – 11) = 20.
 After getting the result of second term we add the result in 23 so first we add both left hand side digits which are 3 and 0 so 3 + 0 = 3. And after that we add second most digits of both Numbers which are 2 and 2, so 2 + 2 = 4.  Now we write these digits in order, so the final result is
23 + 20 = 40.

To solve that expression first we solve exponent terms 5² and 3². So
=> 5² = 25, 3² = 9.
After that we solve multiplication between 5² and 3^2.
=> 25 * 9 = 225,
So the final result of expression 5² * 3² is 225.

In that expression we can clearly see that it contains parenthesis, which have to be solve first.
So the value in parenthesis is 2 and 5 and middle of them there is a multiplication operator. So after solving parenthesis term (2* 5).
(2 * 5) = 10
Now we move from left to right we find that there is a exponent term which is 3⁴ so 3⁴ = 81.
Again moving from left to right we find multiplication and division operators . So first we solve multiplication after that we solve division so
7 + 10 * 81 / 9,
7 + 810 / 9,
7 + 90,
97,
So the final result of the expression 7 + ( 2 * 5) * 3⁴ / 9 is 97.

In that example we can clearly see that it contains addition and division. Now from moving left to right we find division between 36 and 3² so first we solve that so after solving that
=> 36 / 3² = 36 / 9 = 4,
Again moving from left to right now we find addition between 18 and the resulting value of the term (36 / 3²) which is 4. So the resulting answer of the expression 16 + 36 / 3² is:
=> 16 + 4 = 20.

In that example we can clearly see that it contains addition, division and parenthesis. So we have to solve that step by step. From moving left to right we found that there is a parenthesis so we first solve that in whole expression. But here is two parenthesis .so first we solve that which comes first from left to right after that second one.
                                                = (30 -10) / (10 + 10),
                                                = 20 / 20,
                                                = 1.
 So the resulting value of expression (30 -10) / (10 + 10) is 1.

To solve that we have to use order of operation rules. In that example we can clearly see that it contains addition, subtraction, division, multiplication and parenthesis. So we have to solve that step by step. From moving left to right we found that there is a parenthesis so we first solve that in whole expression.
In that parenthesis term there are two operator addition and multiplication so we move from left to right and we find that multiplication is solve first so
                                                150 / (5 + 10) – 5,
Now we solve addition in the parenthesis which is in between 5 and 10 and the result is 10. So
                                                150 / 15 – 5,
Next we again move from left to right in the expression and found that there is division operator which we have to solve then,
                                                10 – 5,
So the result of division is 10. At last we can clearly see that there is only subtraction operator is left in between two values 10 and 5 so we solve that,
                                                10 – 5 = 5,
So the resulting value of expression 150 / ( 5 + 5 * 2) – 5  is 5.