Luca has 13 more crayons than Sam. If Luca has 42 crayons, how many crayons does Sam have ?

Mathematics

1 answer:

Arada [10]2 years ago

5 0

Answer:

<em>Sam has 29 crayons</em>

Step-by-step explanation:

This problem does not really need to set up an equation to solve. We'll give the answer by simple logic deduction and simple calculations.

We know Luca has 13 more crayons than Sam. We can say that Sam has 13 less crayons than Luca.

Now we are told that Luca has 42 crayons. Considering the above relation found for the number of crayons Sam has, we can assure Sam has 42-13=29 crayons.

Conclusion: Sam has 29 crayons

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5-4. Solve each of the following equations for the indicated variable. Use your Equation Mat if it is helpful. Write down each o

balu736 [363]

QUESTION 1

The given equation is :

$2(y - 3) = 4$
We want to solve for y. This means we have to isolate y on one side of the equation while all other constants are also on the other side of the equation,

We first divide both sides by 2 to obtain,

$\frac{2(y - 3)}{2} = \frac{4}{2}$

We cancel out the common factors to get,

$y - 3 = 2$

We now group like terms to get,

$y = 3 + 2$

We simplify to obtain,

$y = 5$

QUESTION 2

The given equation is

$2x + 5y = 10$
We want to solve for x. This means that we need to isolate x on one side of the equation, while all other variables and constant are also on the other side of the equation.

This implies that,

$2x= 10 - 5y$

We now divide through by 2, to obtain,

$\frac{2x}{2} = \frac{10}{2} - \frac{5y}{2}$

This will now give us,

$x = 5 - \frac{5}{2} y$

QUESTION 3

The given equation is
$6x + 3y = 4y + 11$

We want to solve for y. This means that we need to isolate y on one side of the equation, while all other variables and constants are also on the other side of the equation.

We first of all group all the y terms on one side of the equation to obtain,

$6x - 11 = 4y - 3y$

We simplify to get,

$6x - 11 = y$

This implies that,

$y = 6x - 11$

QUESTION 4

The given equation is,

$3(2x + 4) = 2 + 6x + 10$

We want to solve for x. This means that we need to isolate x on one side of the equation, while all other variables and constant are also on the other side of the equation.

Let us first expand the brackets to get,

$6x + 12 = 2 + 6x + 10$
This implies that

$6x + 12 = 6x + 12$

We group like terms to get,

$6x - 6x = 12 - 12$

We now simplify both sides to get,

$0x = 0$

We don't want x to vanish, so let us try to divide both sides by zero to get,

$x = \frac{0}{0}$

This is an indeterminate form, which implies that, x has infinitely many solutions. This means that all the real numbers are solution to the equation.

$\therefore \: x \in \: R$

QUESTION 5

The given equation is,

$y = - 3x + 6$

We want to solve for x, so we add the additive inverse of 6, which is -6 to both sides of the equation to get,

$y - 6 = - 3x$

We rewrite this to obtain,

$- 3x = y - 6$

We divide through by -3 to get,

$\frac{ - 3x}{ - 3} = \frac{y}{ - 3} - \frac{6}{ - 3}$

This will give us,

$x = - \frac{y}{3} + 2$

or

$x= 2 - \frac{1}{3} y$