Question

Please solve these with the full explained answer

Answer 1

Answer:

The given equation is

4x+10=-26

We add the additive inverse of 10 which is -10 to both sides of the equation to obtain,

4x+10+-10=-26+-10

This gives us,

4x+10-10=-26-10

4x+0=-36

Answer 2

ANSWER TO QUESTION 1

The given equation is

$4x+10=-26$

We add the additive inverse of $10$ which is $-10$ to both sides of the equation to obtain,

$4x+10+-10=-26+-10$

This gives us,

$4x+10-10=-26-10$

$4x+0=-36$

This further simplifies to  

$4x=-36$

We divide both sides $4$ to obtain,

$x=-4$

ANSWER TO QUESTION 2

Since we got $-4$ as answer to question one, the next question to answer is

$\frac{x}{3}+10=15$

We add the additive inverse of $10$ which is $-10$ to both sides to obtain,

$\frac{x}{3}+10+-10=15+-10$

This gives us,

$\frac{x}{3}+10-10=15-10$

This simplifies to

$\frac{x}{3}+0=5$

$\frac{x}{3}=5$

We now multiply both sides of the equation by $3$ to obtain,

$3\times \frac{x}{3}=5\times3$

This implies that,

$x=15$

ANSWER TO QUESTION 3

The next equation is $9-2x=35$

We add the additive inverse of $9$ which is $-9$ to both sides of the equation to obtain,

$9+-9-2x=35+-9$

This gives us,

$9-9-2x=35-9$

This simplifies to,

$-2x=26$

We multiply both sides by the reciprocal or the multiplicative inverse of $-2$ to obtain,

$-2x\times (-\frac{1}{2})=26\times(-\frac{1}{2})$

This gives us,

$x=-13$

ANSWER TO QUESTION 4.

The next question is $\frac{2}{3}x+15=17$

We add the additive inverse of 15 to both sides of the equation to obtain,

$\frac{2}{3}x+15+-15=17+-15$

This evaluates to

$\frac{2}{3}x+0=17-15$

This implies that,

$\frac{2}{3}x=2$

We multiply both sides by the multiplicative inverse or the reciprocal of $\frac{2}{3}$

This implies that,

$x=2\times\frac{3}{2}$

$x=3$

ANSWER TO QUESTION 5

The next question to answer is $-5x-10=10$.

We add the additive inverse of $-10$ which is $10$ to both sides of the equation to obtain,

$-5x-10+10=10+10$

This gives us,

$-5x=20$

We through by $-10$ to obtain,

$x=-4$

ANSWER TO QUESTION 6

$\frac{3}{4}x-9=27$

We add the additive inverse of $-9$ which is $9$ to both sides of the equation to obtain,

$\frac{3}{4}x-9+9=27+9$

This simplifies to;

$\frac{3}{4}x=36$

We multiply both sides of the equation by the reciprocal of $\frac{3}{4}$ to obtain,

$\frac{3}{4}x \times \frac{4}{3}=36\times \frac{4}{3}$

$x=36\times \frac{4}{3}$

$x=12\times 4$

$x=48$

ANSWER TO QUESTION 7

The next question to answer is $\frac{1}{2}x+13=9$.

We add the additive inverse of $13$ which is $-13$ to both sides of the equation to obtain,

$\frac{1}{2}x+13+-13=9+-13$

This will evaluate to,

$\frac{1}{2}x=-4$

We now multiply both sides of the equation by $2$ to obtain,

$x=-4\times 2$

$x=-8$

ANSWER TO QUESTION 8

The last question to solve now is $\frac{x-7}{4}=-2$

We first of all multiply both sides of the equation by $4$ to obtain,

$\frac{x-7}{4} \times 4=-2\times 4$

This implies that,

$x-7=-8$

We now add the additive inverse of $-7$ which is 7 to both sides of the equation to obtain,

$x-7+7=-8+7$

This implies that

$x=-1$