You need to solve this equation for x: ax−7=26
Which would be the best first step you could make to solve the equation?
add 7 to both sides of the equation
add 7 to both sides of the equation
subtract 7 from the left side of the equation
subtract 7 from the left side of the equation
Combine ax - 7 to get -7x
Combine ax - 7 to get -7x
divide ax by a and 26 by a
divide ax by a and 26 by a
The best first step to solve the equation would be to add 7 to both sides of the equation.
5x−12−6x=−10x+7−1
step by step
First, let's simplify both sides of the equation by combining like terms:
5x - 12 - 6x = -10x + 7 - 1
Combine the x terms on the left side: (5x - 6x = -x)
-x - 12 = -10x + 7 - 1
Combine the constants on the right side: (7 - 1 = 6)
-x - 12 = -10x + 6
Next, let's isolate the variables on one side of the equation. In this case, let's isolate the -x term on the left side by adding 10x to both sides:
-x + 10x - 12 = -10x + 10x + 6
Combine the x terms on the left side: (-x + 10x = 9x)
9x - 12 = 6
Now, let's isolate the constant term on the right side by adding 12 to both sides:
9x - 12 + 12 = 6 + 12
Combine the constants on the left side: (-12 + 12 = 0)
9x = 18
Finally, let's solve for x by dividing both sides of the equation by 9:
9x/9 = 18/9
x = 2
Therefore, the solution to the equation is x = 2.
The number of x's on both sides of the equation is different a≠b
I apologize for the confusion in my previous response. If the number of x's on both sides of the equation is different (a ≠ b), we need to adjust our steps. Here's the correct step-by-step solution:
5x - 12 - 6x = -10x + 7 - 1
Combine the x terms on the left side: (5x - 6x = -x)
-x - 12 = -10x + 6
Now, let's isolate the x terms on one side of the equation. In this case, let's isolate the -x term on the left side by adding 10x to both sides:
-x + 10x - 12 = -10x + 10x + 6
Combine the x terms on the left side: (-x + 10x = 9x)
9x - 12 = 6
Next, let's isolate the constant term on the left side by adding 12 to both sides:
9x - 12 + 12 = 6 + 12
Combine the constants on the left side: (-12 + 12 = 0)
9x = 18
Finally, let's solve for x by dividing both sides of the equation by 9:
9x/9 = 18/9
x = 2
Therefore, the solution to the equation is x = 2.
ax+c1=bx+c2
The number of x's on both sides of the equation is different a≠b
To solve the equation ax + c1 = bx + c2 when a ≠ b, we need to isolate the x term on one side of the equation. Here's the step-by-step solution:
Step 1: Subtract bx from both sides of the equation to isolate the x term on the left side:
ax + c1 - bx = bx + c2 - bx
This simplifies to:
ax - bx + c1 = c2
Step 2: Factor out x from the terms on the left side:
x(a - b) + c1 = c2
Step 3: Divide both sides of the equation by (a - b) to solve for x:
x = (c2 - c1)/(a - b)
Therefore, the solution to the equation ax + c1 = bx + c2 when a ≠ b is x = (c2 - c1)/(a - b).
The number of x's is the same on both sides of the equation, and the constant is different a=b, c1≠c2
If the number of x's is the same on both sides of the equation (a = b) but the constants are different (c1 ≠ c2), we can solve the equation by isolating the x term. Here's the step-by-step solution:
Step 1: Subtract c1 from both sides of the equation to isolate the x term on the left side:
ax + c1 - c1 = bx + c2 - c1
Simplifying the equation, we get:
ax = bx + (c2 - c1)
Step 2: Subtract bx from both sides of the equation to gather the x term on the left side:
ax - bx = (c2 - c1)
Factoring out x on the left side, we have:
x(a - b) = (c2 - c1)
Step 3: Divide both sides of the equation by (a - b) to solve for x:
x = (c2 - c1)/(a - b)
Therefore, the solution to the equation ax + c1 = bx + c2 when a = b and c1 ≠ c2 is x = (c2 - c1)/(a - b).
Match the description of the one variable equation with the number of solutions it will have.
ax+c1=bx+c2(3 points)
The number of x's on both sides of the equation is different a≠b
The number of x's is the same on both sides of the equation, and the constant is different a=b, c1≠c2
The number of x's is the same on both sides of the equation and the constant is also the same a=b, c1=c2