Let's solve each equation step by step:
1) 9(k+1/3)+10(1/2k-1)=14k-7
Step 1: Distribute the numbers outside the parentheses:
9k + 9/3 + 10/2k - 10 = 14k - 7
Step 2: Combine the like terms:
9k + 3 + 5/k - 10 = 14k - 7
Step 3: Combine the fractions with a common denominator:
9k - 7k/1k + 3 = 14k - 7
Step 4: Combine the like terms:
2k + 3 = 14k - 7
Step 5: Move the variable terms to one side and the constant terms to the other side:
2k - 14k = -7 - 3
Step 6: Combine the like terms:
-12k = -10
Step 7: Solve for k by dividing both sides by -12:
k = -10 / -12
k = 5/6
Therefore, the solution is k = 5/6.
2) 4(g-2)+(g+2)=5g-6
Step 1: Distribute the numbers outside the parentheses:
4g - 8 + g + 2 = 5g -6
Step 2: Combine the like terms:
5g - 6 = 5g - 6
Step 3: The equation is already in its simplified form.
There are infinitely many solutions for this equation.
3) (h+k)-(2k-5) = 2h-k
Step 1: Distribute the negative sign to the terms inside the parentheses:
h + k - 2k + 5 = 2h - k
Step 2: Combine the like terms:
h - k + 5 = 2h - k
Step 3: Move all the variable terms to one side and the constant terms to the other side:
h - 2h = k - k - 5
Step 4: Combine the like terms:
-h = -5
Step 5: Solve for h by multiplying both sides by -1:
h = 5
Therefore, the solution is h = 5.
4) (p-3)-(2.4+2p) = -p - 5.4
Step 1: Simplify the equation by performing the arithmetic inside the parentheses:
p - 3 - 2.4 - 2p = -p - 5.4
Step 2: Combine the like terms:
-p - 5.4 = -p - 5.4
Step 3: The equation is already in its simplified form.
There are infinitely many solutions for this equation.
5) 2(k-4) - 3(2k-1) = -4k - 5
Step 1: Distribute the numbers outside the parentheses:
2k - 8 - 6k + 3 = -4k - 5
Step 2: Combine the like terms:
-4k - 5 = -4k - 5
Step 3: The equation is already in its simplified form.
There are infinitely many solutions for this equation.
7th grade math add and subtract algebraic expressions
9(k+1/3)+10(1/2k-1)=14k-7
4(g-2)+(g+2)=5g-6
(h+k)-(2k-5)=2h-k
(p-3)-(2.4+2p)=-p-5.4
2(k-4)-3(2k-1)=-4k-5
1 answer
1 answer