Let the total toys be x.
The first has (1/10)x = x/10
The third child has one more toys than the first = x/10 + 1
The fourth has double of the third: = 2* (x/10 + 1) = 2x/10 + 2
Since we know that the second child has 12 more toys than the first
Therefore the Second minus First = 12.
First let us find the second.
1st + 2nd + 3rd + 4th = x
x/10 + 2nd + (x/10 + 1) + (2x/10 + 2) = x
2nd + x/10 + x/10 + 1 + 2x/10 + 2 = x
2nd + x/10 + x/10 + 2x/10 + 1 + 2 = x
2nd + 4x/10 + 3 = x
2nd = x - 4x/10 - 3
2nd = 6x/10 - 3
But recall
2nd - 1st = 12
6x/10 - 3 - x/10 = 12
6x/10 - x/10 = 12 + 3
5x/10 = 15
5x = 15 * 10
x = 15*10 /5
x = 30
So there were 30 toys in all.
I would love to use substitution (It is simplier that way I guess)
solve your system of equations.
x+2y=−1;x−y=5
Solve x+2y=−1 for x:
x+2y+−2y=−1+−2y(Add -2y to both sides)
x=−2y−1
Substitute (−2y−1) for x in x−y=5:
x−y=5
−2y−1−y=5
−3y−1=5(Simplify both sides of the equation)
−3y−1+1=5+1(Add 1 to both sides)
−3y=6
−3y/−3 = 6/−3(Divide both sides by -3)
y=−2
Substitute (−2) for y in x=−2y−1:
x=−2y−1
x=(−2)(−2)−1
x=3(Simplify both sides of the equation)
So the answer is (x = 3 and y = -2)
Answer:
45
Step-by-step explanation:
I think so i dont know
You find the least common multiple of the two denominators, then you multiply whatever you multiplied to the denominator to the numerator to get to the LCM.
ex:
3/4 + 2/3 --> 9/12+ 8/12= 17/12
remember, you can do anything to an equation as long as you do it to both sides
try and isolate the variable you are solving for and divide it by its constant
p=2l+2w
minus 2l from both sides
p-2l=2w
divide both sides by 2
which can also be simplified to
