QUESTION 1
The given equation is :
[tex]2(y - 3) = 4[/tex]
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,
[tex] \frac{2(y - 3)}{2} = \frac{4}{2} [/tex]
We cancel out the common factors to get,
[tex]y - 3 = 2[/tex]
We now group like terms to get,
[tex]y = 3 + 2[/tex]
We simplify to obtain,
[tex]y = 5[/tex]
QUESTION 2
The given equation is
[tex]2x + 5y = 10[/tex]
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,
[tex]2x= 10 - 5y[/tex]
We now divide through by 2, to obtain,
[tex] \frac{2x}{2} = \frac{10}{2} - \frac{5y}{2} [/tex]
This will now give us,
[tex]x = 5 - \frac{5}{2} y[/tex]
QUESTION 3
The given equation is
[tex]6x + 3y = 4y + 11[/tex]
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,
[tex]6x - 11 = 4y - 3y[/tex]
We simplify to get,
[tex]6x - 11 = y[/tex]
This implies that,
[tex]y = 6x - 11[/tex]
QUESTION 4
The given equation is,
[tex]3(2x + 4) = 2 + 6x + 10[/tex]
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,
[tex]6x + 12 = 2 + 6x + 10[/tex]
This implies that
[tex]6x + 12 = 6x + 12[/tex]
We group like terms to get,
[tex]6x - 6x = 12 - 12[/tex]
We now simplify both sides to get,
[tex]0x = 0[/tex]
We don't want x to vanish, so let us try to divide both sides by zero to get,
[tex]x = \frac{0}{0} [/tex]
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.
[tex] \therefore \: x \in \: R[/tex]
QUESTION 5
The given equation is,
[tex]y = - 3x + 6[/tex]
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,
[tex]y - 6 = - 3x[/tex]
We rewrite this to obtain,
[tex] - 3x = y - 6[/tex]
We divide through by -3 to get,
[tex] \frac{ - 3x}{ - 3} = \frac{y}{ - 3} - \frac{6}{ - 3} [/tex]
This will give us,
[tex]x = - \frac{y}{3} + 2[/tex]
or
[tex]x= 2 - \frac{1}{3} y[/tex]
