That's an extraneous solution. You need to check for these when algebraically solving equations, especially when you take both sides of an equation to a power.
-13x = 52 x = -4 Check: -13*-4 = 52
It depends whether it is a subtraction, mulitplication, addition, or division problem.
Set up algebraically. X = what X/2769 = 5 2769/1*(X/2769 = 5)*2769/1 multiply through by 2769 X = 13845 ------------- check 13845/2769 = 5 5 = 5 checks
yes. check: 3x-6=5x-10 2x=4. X=2. yay!
That's an extraneous solution. You need to check for these when algebraically solving equations, especially when you take both sides of an equation to a power.
Algebraically. 24X = 72 (24/24)X = 72/24 X = 3 --------------check 24(3) = 72 72 = 72 ----------------checks
Check your text book for how to solve it.
-13x = 52 x = -4 Check: -13*-4 = 52
It depends whether it is a subtraction, mulitplication, addition, or division problem.
(Since 5 times 20 does equal 100, the solution is correct.
No, it is not. Substitute and check. You have 128 + 16r. So now, if r=9, then the equation would be 128 = 16x9. But, 16x9 is 145 and is not equal to 128. Thus, r=9 is not the solution. The solution is r+ 128/16 which is equal to 8. Thus, the actual solution is r = 8.
Check ouhttp://hspm.sph.sc.edu/COURSES/ECON/Elast/Elast.htmlt on
Set up algebraically. X = what X/2769 = 5 2769/1*(X/2769 = 5)*2769/1 multiply through by 2769 X = 13845 ------------- check 13845/2769 = 5 5 = 5 checks
You write the equation in such a way that you have zero on the right side. Then you graph the expression on the left side of the equal sign, and check where it touches the x-axis. Note that this method works for most common equations.
yes. check: 3x-6=5x-10 2x=4. X=2. yay!
You can check an answer by substituting it into the equation. x - 17 = 11 25 - 17 = 11 8 = 11 Since this is not true, 25 is not the answer.