ANSWER TO QUESTION 1
The given equation is
[tex]4x+10=-26[/tex]
We add the additive inverse of [tex]10[/tex] which is [tex]-10[/tex] to both sides of the equation to obtain,
[tex]4x+10+-10=-26+-10[/tex]
This gives us,
[tex]4x+10-10=-26-10[/tex]
[tex]4x+0=-36[/tex]
This further simplifies to
[tex]4x=-36[/tex]
We divide both sides [tex]4[/tex] to obtain,
[tex]x=-4[/tex]
ANSWER TO QUESTION 2
Since we got [tex]-4[/tex] as answer to question one, the next question to answer is
[tex]\frac{x}{3}+10=15[/tex]
We add the additive inverse of [tex]10[/tex] which is [tex]-10[/tex] to both sides to obtain,
[tex]\frac{x}{3}+10+-10=15+-10[/tex]
This gives us,
[tex]\frac{x}{3}+10-10=15-10[/tex]
This simplifies to
[tex]\frac{x}{3}+0=5[/tex]
[tex]\frac{x}{3}=5[/tex]
We now multiply both sides of the equation by [tex]3[/tex] to obtain,
[tex]3\times \frac{x}{3}=5\times3[/tex]
This implies that,
[tex]x=15[/tex]
ANSWER TO QUESTION 3
The next equation is [tex]9-2x=35[/tex]
We add the additive inverse of [tex]9[/tex] which is [tex]-9[/tex] to both sides of the equation to obtain,
[tex]9+-9-2x=35+-9[/tex]
This gives us,
[tex]9-9-2x=35-9[/tex]
This simplifies to,
[tex]-2x=26[/tex]
We multiply both sides by the reciprocal or the multiplicative inverse of [tex]-2[/tex] to obtain,
[tex]-2x\times (-\frac{1}{2})=26\times(-\frac{1}{2})[/tex]
This gives us,
[tex]x=-13[/tex]
ANSWER TO QUESTION 4.
The next question is [tex]\frac{2}{3}x+15=17[/tex]
We add the additive inverse of 15 to both sides of the equation to obtain,
[tex]\frac{2}{3}x+15+-15=17+-15[/tex]
This evaluates to
[tex]\frac{2}{3}x+0=17-15[/tex]
This implies that,
[tex]\frac{2}{3}x=2[/tex]
We multiply both sides by the multiplicative inverse or the reciprocal of [tex]\frac{2}{3}[/tex]
This implies that,
[tex]x=2\times\frac{3}{2}[/tex]
[tex]x=3[/tex]
ANSWER TO QUESTION 5
The next question to answer is [tex]-5x-10=10[/tex].
We add the additive inverse of [tex]-10[/tex] which is [tex]10[/tex] to both sides of the equation to obtain,
[tex]-5x-10+10=10+10[/tex]
This gives us,
[tex]-5x=20[/tex]
We through by [tex]-10[/tex] to obtain,
[tex]x=-4[/tex]
ANSWER TO QUESTION 6
[tex]\frac{3}{4}x-9=27[/tex]
We add the additive inverse of [tex]-9[/tex] which is [tex]9[/tex] to both sides of the equation to obtain,
[tex]\frac{3}{4}x-9+9=27+9[/tex]
This simplifies to;
[tex]\frac{3}{4}x=36[/tex]
We multiply both sides of the equation by the reciprocal of [tex]\frac{3}{4}[/tex] to obtain,
[tex]\frac{3}{4}x \times \frac{4}{3}=36\times \frac{4}{3}[/tex]
[tex]x=36\times \frac{4}{3}[/tex]
[tex]x=12\times 4[/tex]
[tex]x=48[/tex]
ANSWER TO QUESTION 7
The next question to answer is [tex]\frac{1}{2}x+13=9[/tex].
We add the additive inverse of [tex]13[/tex] which is [tex]-13[/tex] to both sides of the equation to obtain,
[tex]\frac{1}{2}x+13+-13=9+-13[/tex]
This will evaluate to,
[tex]\frac{1}{2}x=-4[/tex]
We now multiply both sides of the equation by [tex]2[/tex] to obtain,
[tex]x=-4\times 2[/tex]
[tex]x=-8[/tex]
ANSWER TO QUESTION 8
The last question to solve now is [tex]\frac{x-7}{4}=-2[/tex]
We first of all multiply both sides of the equation by [tex]4[/tex] to obtain,
[tex]\frac{x-7}{4} \times 4=-2\times 4[/tex]
This implies that,
[tex]x-7=-8[/tex]
We now add the additive inverse of [tex]-7[/tex] which is 7 to both sides of the equation to obtain,
[tex]x-7+7=-8+7[/tex]
This implies that
[tex]x=-1[/tex]
