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k=4

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Q: What is 3k plus 5 equals 2k plus 1?
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3K-1 equals K plus 2?

3k-1=k+2 2k=3 k=3/2=1.5


How do you solve 3k squared plus 2k-5?

When factored it is: (3k+5)(k-1)


How do you find the product for this problem 3kminus6 times2k plus 1?

The question is unclear, so the author will provide answers for a number of interpretations: 1. 3k-6(2k+1) = 3k-12k-6=-9k-6=-3(3k+2) 2. 3k-6(2k)+1=3k-12k+1=-9k+1 3. (3k-6)(2k)+1 = 6k^2 -12k + 1 = 6(k-1-sqrt(5/6))(k-1+sqrt(5/6)) 4. (3k-6)(2k+1) = 6k^2 - 12k + 3k - 6 = 6k^2 -9k + 6 = 3(2k^2 - 3k + 2) Line 4 cannot be factorised further. sqrt and ^2 refer to the square root, and squared respectively. Lines 1 and 2 require knowledge of expansion of linear equations, addition of like terms, and factorisation of linear equations. Lines 3 and 4 also require knowledge of addition of like terms, and expansion and factorisation of quadratic equations. In no case can an exact value for k be determined as we were given an expression rather than an equality.


Combine the like terms to create an equivalent expression:5k+(−2k)−(−1)?

3k + 1


If 2k-1 equals 0 then k equals?

2k - 1 = 0 Add 1 to both sides: 2k = 1 Divide both sides by two: k = 0.5


K plus 1 equals 3k - 1?

K+1 is as simplified as it gets. Unless you find the value of K. You can't add a variable and number.


What property is 1 plus 3k plus 6?

It is the associative property of addition in the sense that a sum of three elements is unambiguous even if given without brackets to indicate which of the sums is carried out first. This would not apply for subtraction since 1 - (3k - 6) = 1 - 3k + 6 whereas (1 - 3k) - 6 = 1 - 3k - 6


Can someone prove that 2 raised to the k equals 2 raised to the k plus 1 minus 1 by induction?

Is the following what you are claiming? 2k = 2k+1 -1 20 not equal 20+1 - 1 21 not equal 21+1 -1


What is k plus 1 plus k plus 4?

2k + 5


What does odd X even equal?

Let even be of the form 2k and odd be of the form 2k+1. Then odd * even becomes 2k*2k+1, or 4k^2 +2k. This can be written as 2(k^2 + k), which is of the form 2k. Therefore, odd X even equals even.


Can someone prove that 2k2k plus 1-1 by induction?

To prove that 2k 2k plus 1-1 by induction is a step by step process. But the induction 2 is not equal to 2 to the power of 0 take away 1.


What are the possible values of k when 2kx2 -2x2 plus 2kx plus k -1 equals 0 has equal roots?

Equation: 2kx^2 -2x^2 +2kx +k -1 = 0 Using the discriminant: (2k)^2 -4*(2k -2)*(k -1) = 0 Solving for k in the discriminant: k = 2 + or - square root of 2