(x) (x 1) (x 2)=111 3x 1 2=111 3x 3=111 If you wanted to solve for x, you would isolate the variable by subtracting 3 from both sides, then dividing by 3: 3x 3=111 3x=108 x=36 So, the 3 consecutive integers are 36, 37, and 38. A This problem involves translating the words into mathematical expressions.
The first equation is quite simple to set up, since you know that she works 5 hours each week: x y=5 Use the information about the money Audrey earns to set up the second equation.
So his normal pay of 40 × $12 = $480, plus his overtime pay of 12 × $15 = $180 gives us a total of $660 There are 12 girls!
And 3b = 4g, so b = 4g/3 = 4 × 12 / 3 = 16, so there are 16 boys So there are now 12 girls and 16 boys in the class, making 28 students altogether.
Things get a little more complicated in algebra when it comes to word problems, however.
Sometimes, you’ll be given a word problem that requires you to WRITE the equation first and THEN solve it. C) She tutors for 2 hours and babysits for 3 hours. She sells necklaces for , bracelets for , and rings for . How many necklaces, bracelets, and rings did Iris sell?
Check −14: −14(−14 2) = (−14)×(−12) = 168 YES Check 12: 12(12 2) = 12×14 = 168 YES So there are two solutions: -14 and -12 is one, 12 and 14 is the other.
Note: we could have also tried "guess and check": And so L = 8 or −14 There are two solutions to the quadratic equation, but only one of them is possible since the length of the room cannot be negative!
This should result in the following equations: K= ¾ Z, S = ⅔ Z, K = S 15Taking the equation K=S 15 students need to replace the K with K= ¾ Z and the S with S= ⅔ Z thus creating the equation ¾ Z= ⅔ Z 15Students then need to recognise and multiply each of the terms by the common denominator 12, and then solve the resulting equation 9Z= 8Z 15 ∴ Z = 15Students should first review the process of writing algebraic expressions and equations from a worded description or rule.
The resource discusses and explains determining a formula to reinforce students' understanding.