Answer:
[tex]1)\ \ 4h^2-13h+6\\2)\ \ 7x^3y^2-x^2y+1\\3)\ \ -7n+2\\4)\ \ -8m+4[/tex]
Step-by-step explanation:
1.
Simplify the expression by combining like terms. Remember, like terms have the same variable part, to simplify these terms, one performs operations between the coefficients. Please note that a variable with an exponent is not the same as a variable without the exponent. A term with no variable part is referred to as a constant, constants are like terms.
[tex]2h^2-7h+2h^2-h+6+4h-9h[/tex]
[tex](2h^2+2h^2)+(-7h-h+4h-9h)+(6)[/tex]
[tex]4h^2-13h+6[/tex]
2.
Use a very similar method to solve this problem as used in the first. Please note that all of the rules mentioned in the first problem also apply to this problem; for that matter, the rules mentioned in the first problem generally apply to any pre-algebra problem.
[tex]8x^3y^2-7x^2y+8x-4-x^3y^2+2x^2y+4x^2y-8x+5[/tex]
[tex](8x^3y^2-x^3y^2)+(-7x^2y+2x^2y+4x^2y)+(8x-8x)+(-4+5)[/tex]
[tex]7x^3y^2-x^2y+1[/tex]
3.
Use the same rules as applied in the first problem. Also, keep the distributive property in mind. In simple terms, the distributive property states the following ([tex]a(b+c)=(a)(b)+(a)(c)=ab+ac[/tex]). Also note, a term raised to an exponent is equal to the term times itself the number of times the exponent indicates. In the event that the term raised to an exponent is a constant, one can simplify it. Apply these properties here,
[tex]-2(8n+1)-(5-9n)+3^2[/tex]
[tex]-2(8n+1)-(5-9n)+(3*3)[/tex]
[tex]-2(8n+1)-(5-9n)+9[/tex]
[tex](-2)(8n)+(-2)(1)+(-)(5)+(-)(-9n)+9[/tex]
[tex]-16n-2-5+9n+9[/tex]
[tex](-16n+9n)+(-2-5+9)[/tex]
[tex]-7n+2[/tex]
4.
The same method used to solve problem (3) can be applied to this problem.
[tex]\frac{1}{2}(10-8m+6m^2)-(3m^2+4m-7)-2^3[/tex]
[tex]\frac{1}{2}(10-8m+6m^2)-(3m^2+4m-7)-(2)(2)(2)[/tex]
[tex](\frac{1}{2})(10)+(\frac{1}{2})(-8m)+(\frac{1}{2})(6m^2})+(-)(3m^2)+(-1)(4m)+(-1)(-7)-8[/tex]
[tex]5-4m+3m^2-3m^2-4m+7-8[/tex]
[tex](-3m^2-3m^2)+(-4m-4m)+(5+7-8)[/tex]
[tex]-8m+4[/tex]
