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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

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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

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