Solving Equations by using Addition/ Subtraction: Solving an equation using all techniques

$4(\frac{b}{2}+b)-\frac{1}{3}=6b$

$b=\text{?}$

First, we'll open the parentheses by multiplying each term by 4:

$4\times\frac{b}{2}+4\times b-\frac{1}{3}=6b$

Let's solve the multiplication exercise $4\times\frac{b}{2}=\frac{4b}{2}=2b$

Now the equation is:

$2b+4b-\frac{1}{3}=6b$

We'll combine the left side between the two b terms and get:

$6b-\frac{1}{3}=6b$

We'll reduce both sides by 6b and get:

$-\frac{1}{3}=0$

Since the result obtained is impossible, the exercise has no solution.

No solution

$37b+6b+56=90+9$

$b=\text{?}$

1

Solve for X:

$8+2(x+1)=7x$

$2$

$\frac{1}{4}a+5=20+a$

$a=\text{?}$

$-20$

$6c+7+4c=3(c-1)$

$c=\text{?}$

$-1\frac{3}{7}$

$4y-7+6y=3-10y$

$y=?$

$\frac{1}{2}$

Solve for X:

$3-(x-4)+8x=14$

$1$

$7y+10y+5=2(y+3)$

$y=\text{?}$

$\frac{1}{15}$

$16a-20a+15=2(5-2a)$

$a=\text{?}$

No solution

$-3(4a+8)=27a$

$a=\text{?}$

$-\frac{8}{13}$

$\frac{1}{3}(x+9)=4+\frac{2}{3}x$

$x=\text{?}$

3-

Solve for X:

$(3-2x)\cdot5=4+12x$

$\frac{1}{2}$

$a^4+7a-5=2a+a^4+3a-(-a)$

$a=?$

$5$

Solve for X:

$6\cdot(5+2x)=3x-6$

$-4$

$12y+3y-10+7(y-4)=2y$

$y=?$

$1.9$

$3x+\frac{2}{3}=4(x+\frac{1}{12})$

$x=\text{?}$

$\frac{1}{3}$

Solve for X:

$\frac{1}{4}(x-16)+\frac{3}{4}x=8x-11$

$1$

$-\frac{7}{4}(-x)+2x-5(x+3)=-x$

$x=\text{?}$

$-60$

Solve for X:

$\frac{1}{2}\cdot(8-x)+\frac{1}{2}x=12-x$

$8$

$2x+45-\frac{1}{3}x=5(x+7)$

$x=\text{?}$

3