Given:The given equation is $\frac{2}{3}(x-5)-\frac{1}{4}(x-2)=\frac{9}{2}$.To do:We have to solve the given equation and verify the solution.Solution:To verify the solution we have to find the value of the variable and substitute it in the equation. Find the value of LHS and the value of RHS and check whether both are equal.The given equation is $\frac{2}{3}(x-5)-\frac{1}{4}(x-2)=\frac{9}{2}$.$\frac{2(x-5)}{3}-\frac{1(x-2)}{4}=\frac{9}{2}$$\frac{2x-10}{3}-\frac{x-2}{4}=\frac{9}{2}$LCM of denominators $3$ and $4$ is $12$.$\frac{(2x-10)\times4-(x-2)\times3}{12}=\frac{9}{2}$$\frac{8x-40-3x+6}{12}=\frac{9}{2}$$\frac{5x-34}{12}=\frac{9}{2}$On cross multiplication, we get, $(5x-34=\frac{9\times12}{2}$$5x-34=9\times6$$5x-34=54$$5x=54+34$$5x=88$$x=\frac{88}{5}$Verification:LHS $=\frac{2}{3}(x-5)-\frac{1}{4}(x-2)$$=\frac{2}{3}(\frac{88}{5}-5)-\frac{1}{4}(\frac{88}{5}-2)$$=\frac{2}{3}(\frac{88-5\times5}{5})-\frac{1}{4}(\frac{88-2\times5}{5})$$=\frac{2}{3}(\frac{88-25}{5})-\frac{1}{4}(\frac{88-10}{5})$$=\frac{2}{3}(\frac{63}{5})-\frac{1}{4}(\frac{78}{5})$$=\frac{2}{1}(\frac{21}{5})-\frac{1}{2}(\frac{39}{5})$$=\frac{42}{5}-\frac{39}{10}$$=\frac{42\times2-39}{10}$$=\frac{84-39}{10}$$=\frac{45}{10}$$=\frac{9}{2}$RHS $=\frac{9}{2}$LHS $=$ RHSHence verified.Read More

Given:The given equations are:(i) $\frac{(2x-1)}{3}-\frac{(6x-2)}{5}=\frac{1}{3}$(ii) $13(y-4)-3(y-9)-5(y+4)=0$To do:We have to solve the given equations and verify the solutions.Solution:To verify the solutions we have to find the values of the variables and substitute them in the equation. Find the value of LHS and the value of RHS and check whether both are equal.(i) The given equation is $\frac{(2x-1)}{3}-\frac{(6x-2)}{5}=\frac{1}{3}$.$\frac{(2x-1)}{3}-\frac{(6x-2)}{5}=\frac{1}{3}$LCM of denominators $3$ and $5$ is $15$$\frac{(2x-1)\times5-(6x-2) \times3}{15}=\frac{1}{3}$$\frac{10x-5-18x+6}{15}=\frac{1}{3}$$\frac{-8x+1}{15}=\frac{1}{3}$On cross multiplication, we get, $-8x+1=\frac{1\times15}{3}$$-8x+1=5$$8x=1-5$$8x=-4$$x=\frac{-4}{8}$$x=\frac{-1}{2}$Verification:LHS $=\frac{(2x-1)}{3}-\frac{(6x-2)}{5}$$=\frac{(2\times\frac{-1}{2}-1)}{3}-\frac{(6\times\frac{-1}{2}-2)}{5}$$=\frac{-1-1}{3}-\frac{-3-2}{5}$$=\frac{-2}{3}-\frac{-5}{5}$$=\frac{-2}{3}+1$$=\frac{-2+1\times3}{3}$$=\frac{-2+3}{3}$$=\frac{1}{3}$RHS $=\frac{1}{3}$LHS $=$ RHSHence verified.(ii) The given equation is $13(y-4)-3(y-9)-5(y+4)=0$.$13(y-4)-3(y-9)-5(y+4)=0$$13y-52-3y+27-5y-20=0$$13y-8y-72+27=0$$5y-45=0$$5y=45$$y=\frac{45}{5}$$y=9$Verification:LHS $=13(y-4)-3(y-9)-5(y+4)$$=13(9-4)-3(9-9)-5(9+4)$$=13(5)-3(0)-5(13)$$=65-0-65$$=0$RHS $=0$LHS $=$ RHSHence verified.Read More

Given:The given equations are:(i) $\frac{x}{2}-\frac{4}{5}+\frac{x}{5}+\frac{3x}{10}=\frac{1}{5}$(ii) $\frac{7}{x}+35=\frac{1}{10}$To do:We have to solve the given equations and verify the solutions.Solution:To verify the solutions we have to find the values of the variables and substitute them in the equation. Find the value of LHS and the value of RHS and check whether both are equal.(i) The given equation is $\frac{x}{2}-\frac{4}{5}+\frac{x}{5}+\frac{3x}{10}=\frac{1}{5}$.$\frac{x}{2}-\frac{4}{5}+\frac{x}{5}+\frac{3x}{10}=\frac{1}{5}$$\frac{x}{2}+\frac{x}{5}+\frac{3x}{10}=\frac{1}{5}+\frac{4}{5}$     (Transposing $\frac{4}{5}$ to RHS)LCM of denominators $2, 5$ and $10$ is $10$$\frac{x \times5+x \times2+3x \times1}{10}=\frac{1+4}{5}$$\frac{5x+2x+3x}{10}=\frac{5}{5}$$\frac{10x}{10}=1$$x=1$Verification:LHS $=\frac{x}{2}-\frac{4}{5}+\frac{x}{5}+\frac{3x}{10}$$=\frac{1}{2}-\frac{4}{5}+\frac{1}{5}+\frac{3(1)}{10}$$=\frac{1\times5-4\times2+1\times2+3}{10}$    (LCM of $2, 5$ and $10$ is $10$)$=\frac{5-8+2+3}{10}$$=\frac{10-8}{10}$$=\frac{2}{10}$$=\frac{1}{5}$RHS $=\frac{1}{5}$LHS $=$ RHSHence verified.(ii) The given equation is $\frac{7}{x}+35=\frac{1}{10}$.$\frac{7}{x}+35=\frac{1}{10}$$\frac{7}{x}=\frac{1}{10}-35$                    (Transposing $35$ to ... Read More

Given:The given equations are:(i) $\frac{2x}{3}-\frac{3x}{8}=\frac{7}{12}$(ii) $(x+2)(x+3)+(x-3)(x-2)-2x(x+1)=0$To do:We have to solve the given equations and verify the solutions.Solution:To verify the solutions we have to find the values of the variables and substitute them in the equation. Find the value of LHS and the value of RHS and check whether both are equal.(i) The given equation is $\frac{2x}{3}-\frac{3x}{8}=\frac{7}{12}$.$\frac{2x}{3}-\frac{3x}{8}=\frac{7}{12}$LCM of $3$ and $8$ is $24$$\frac{2x \times 8-3x \times3}{24}=\frac{7}{12}$$\frac{16x-9x}{24}=\frac{7}{12}$$\frac{7x}{24}=\frac{7}{12}$On cross multiplication, we get, $7x =\frac{7\times24}{12}$$7x=\frac{7\times2}{1}$$7x=14$$x=\frac{14}{7}$$x=2$Verification:LHS $=\frac{2x}{3}-\frac{3x}{8}$$=\frac{2\times2}{3}-\frac{3\times2}{8}$$=\frac{4}{3}-\frac{3}{4}$$=\frac{4\times4-3\times3}{12}$               (LCM of $3$ and $4$ is $12$)$=\frac{16-9}{12}$$=\frac{7}{12}$RHS $=\frac{7}{12}$LHS $=$ RHSHence verified.(ii) The given equation is $(x+2)(x+3)+(x-3)(x-2)-2x(x+1)=0$.$(x+2)(x+3)+(x-3)(x-2)-2x(x+1)=0$$x(x+3)+2(x+3)+x(x-2)-3(x-2)-2x(x)-2x(1)=0$$x^2+3x+2x+6+x^2-2x-3x+6-2x^2-2x=0$$2x^2-2x^2+5x-7x+12=0$$-2x+12=0$$2x=12$$x=\frac{12}{2}$$x=6$Verification:LHS $=(x+2)(x+3)+(x-3)(x-2)-2x(x+1)$$=(6+2)(6+3)+(6-3)(6-2)-2(6)(6+1)$$=(8)(9)+(3)(4)-12(7)$$=72+12-84$$=84-84$$=0$RHS $=0$LHS $=$ RHSHence verified.Read More

Given:The given equations are:(i) $\frac{x}{2}+\frac{x}{3}+\frac{x}{4}=13$(ii) $\frac{x}{2}+\frac{x}{8}=\frac{1}{8}$To do:We have to solve the given equations and verify the solutions.Solution:To verify the solutions we have to find the values of the variables and substitute them in the equation. Find the value of LHS and the value of RHS and check whether both are equal.(i) The given equation is $\frac{x}{2}+\frac{x}{3}+\frac{x}{4}=13$.$\frac{x}{2}+\frac{x}{3}+\frac{x}{4}=13$LCM of denominators $2, 3$  and $4$ is $12$Therefore, $\frac{x \times6+x \times4+x \times3}{12}=13$$\frac{6x+4x+3x}{12}=13$$\frac{13x}{12}=13$On cross multiplication, we get, $13x=12\times13$$x=\frac{12\times13}{13}$$x=12$Verification:LHS $=\frac{x}{2}+\frac{x}{3}+\frac{x}{4}$$=\frac{12}{2}+\frac{12}{3}+\frac{12}{4}$$=6+4+3$$=13$RHS $=13$LHS $=$ RHSHence verified.(ii) The given equation is $\frac{x}{2}+\frac{x}{8}=\frac{1}{8}$.$\frac{x}{2}+\frac{x}{8}=\frac{1}{8}$LCM of denominators $2$ and $8$ is $8$$\frac{x \times 4+x}{8}=\frac{1}{8}$$\frac{4x+x}{8}=\frac{1}{8}$$\frac{5x}{8}=\frac{1}{8}$On cross multiplication, we get, $5x=\frac{1\times8}{8}$$5x=1$$x=\frac{1}{5}$Verification:LHS $=\frac{x}{2}+\frac{x}{8}$$=\frac{\frac{1}{5}}{2}+\frac{\frac{1}{5}}{8}$$=\frac{1}{5\times2}+\frac{1}{5\times8}$$=\frac{1}{10}+\frac{1}{40}$$=\frac{1\times4+1}{40}$            ... Read More

Given:The given equations are:(i) $9\frac{1}{4}=y-1\frac{1}{3}$(ii) $\frac{5x}{3}+\frac{2}{5}=1$To do:We have to solve the given equations and verify the solutions.Solution:To verify the solutions we have to find the values of the variables and substitute them in the equation. Find the value of LHS and the value of RHS and check whether both are equal.(i) The given equation is $9\frac{1}{4}=y-1\frac{1}{3}$.$9\frac{1}{4}=y-1\frac{1}{3}$$\frac{9\times4+1}{4}=y-\frac{1\times3+1}{3}$$\frac{36+1}{4}=y-\frac{3+1}{3}$$\frac{37}{4}=y-\frac{4}{3}$$y=\frac{37}{4}+\frac{4}{3}$                   (Transposing $\frac{4}{3}$)LCM of the denominators $4$ and $3$ is $12$.$y=\frac{37}{4}+\frac{4}{3}$$y=\frac{37\times3+4\times4}{12}$$y=\frac{111+16}{12}$$y=\frac{127}{12}$Verification:LHS $=9\frac{1}{4}$$=\frac{9\times4+1}{4}$$=\frac{36+1}{4}$$=\frac{37}{4}$RHS $=y-1\frac{1}{3}$$=\frac{127}{12}-1\frac{1}{3}$$=\frac{127}{12}-\frac{1\times3+1}{3}$$=\frac{127}{12}-\frac{3+1}{3}$$=\frac{127}{12}-\frac{4}{3}$$=\frac{127-4\times4}{12}$$=\frac{127-16}{12}$$=\frac{111}{12}$$=\frac{3\times37}{3\times4}$$=\frac{37}{4}$LHS $=$ RHSHence verified.(ii) The given equation is $\frac{5x}{3}+\frac{2}{5}=1$.$\frac{5x}{3}+\frac{2}{5}=1$LCM of denominators $3$ and $5$ is $15$$\frac{5x \times 5+2\times3}{15}=1$$\frac{25x+6}{15}=1$On cross multiplication, we get, $25x+6=15$$25x=15-6$$25x=9$$x=\frac{9}{25}$Verification:LHS $=\frac{5x}{3}+\frac{2}{5}$$=\frac{5\times \frac{9}{25}}{3}+\frac{2}{5}$$=\frac{1\times \frac{3}{5}}{1}+\frac{2}{5}$$=\frac{3}{5}+\frac{2}{5}$$=\frac{3+2}{5}$$=\frac{5}{5}$$=1$RHS $=1$LHS $=$ ... Read More

Given:The given expressions are:(i) $acx^2+(bc+ad)x+bd$ by $ax+b$(ii) $(a^2+2ab+b^2)-(a^2+2ac+c^2)$ by $2a+b+c$To do:We have to divide the given expressions.Solution:We have to divide the given polynomials by simplifying them using algebraic formulas.Polynomials: Polynomials are expressions in which each term is a constant multiplied by a variable raised to a whole number power.Therefore, (i) The given expression is $acx^2+(bc+ad)x+bd$ by $ax+b$.$acx^2+(bc+ad)x+bd \div (ax+b)=\frac{acx^2+(bc+ad)x+bd}{ax+b}$$acx^2+(bc+ad)x+bd \div (ax+b)=\frac{acx^2+bcx+adx+bd}{ax+b}$$acx^2+(bc+ad)x+bd \div (ax+b)=\frac{cx(ax+b)+d(ax+b)}{ax+b}$        (Taking $cx$ and $d$ common)$acx^2+(bc+ad)x+bd \div (ax+b)=\frac{cx(ax+b)}{ax+b}+\frac{d(ax+b)}{ax+b}$$acx^2+(bc+ad)x+bd \div (ax+b)=cx+d$Hence, $acx^2+(bc+ad)x+bd$ divided by $ax+b$ is $cx+d$.(ii) The given expression is $(a^2+2ab+b^2)-(a^2+2ac+c^2)$ by $2a+b+c$.$(a^2+2ab+b^2)-(a^2+2ac+c^2) \div (2a+b+c)=\frac{(a^2+2ab+b^2)-(a^2+2ac+c^2)}{2a+b+c}$$(a^2+2ab+b^2)-(a^2+2ac+c^2) \div (2a+b+c)=\frac{(a+b)^2-(a+c)^2}{2a+b+c}$                [Since $(x+y)^2=x^2+2xy+y^2$]$(a^2+2ab+b^2)-(a^2+2ac+c^2) \div (2a+b+c)=\frac{(a+b+a+c)(a+b-a-c)}{2a+b+c}$      ... Read More

Given:The given expressions are:(i) $ax^2-ay^2$ by $ax+ay$(ii) $x^4-y^4$ by $x^2-y^2$To do:We have to divide the given expressions.Solution:We have to divide the given polynomials by simplifying them using algebraic formulas.Polynomials: Polynomials are expressions in which each term is a constant multiplied by a variable raised to a whole number power.Therefore, (i) The given expression is $ax^2-ay^2$ by $ax+ay$.$ax^2-ay^2$ can be written as, $ax^2-ay^2=a(x^2-y^2)$            (Taking $a$ common)$ax^2-ay^2=a(x+y)(x-y)$.........(I)               [Since $a^2-b^2=(a+b)(a-b)$]Therefore, $ax^2-ay^2 \div (ax+ay)=\frac{ax^2-ay^2}{ax+ay}$$ax^2-ay^2 \div (ax+ay)=\frac{a(x+y)(x-y)}{a(x+y)}$             [Using (I) and taking $a$ common in $ax+ay$]$ax^2-ay^2 \div (ax+ay)=(x-y)$Hence, $ax^2-ay^2$ divided by $ax+ay$ is ... Read More

Given:The given expressions are:(i) $5x^3-15x^2+25x$ by $5x$(ii) $4z^3+6z^2-z$ by $\frac{-1}{2}z$(iii) $9x^2y-6xy+12xy^2$ by $\frac{-3}{2}xy$To do:We have to divide the given expressions.Solution:We have to divide the given polynomials by monomials using the formula $x^a \div x^b=a^{a-b}$Polynomials: Polynomials are expressions in which each term is a constant multiplied by a variable raised to a whole number power.Monomial:A monomial is an expression that contains a single term composed of a product of constants and variables with non-negative integer exponents. Therefore, (i) The given expression is $5x^3-15x^2+25x$ by $5x$.$5x^3-15x^2+25x \div 5x=\frac{5x^3}{5x}-\frac{15x^2}{5x}+\frac{25x}{5x}$$5x^3-15x^2+25x \div 5x=\frac{5}{5}x^{3-1}-\frac{15}{5}x^{2-1}+\frac{25}{5}x^{1-1}$$5x^3-15x^2+25x \div 5x=x^{2}-3x^{1}+5x^{0}$$5x^3-15x^2+25x \div 5x=x^{2}-3x+5$             [Since $x^0=1$]Hence, $5x^3-15x^2+25x$ divided by $5x$ is ... Read More

Given:The given expressions are:(i) $-x^6+2x^4+4x^3+2x^2$ by $\sqrt2x^2$(ii) $-4a^3+4a^2+a$ by $2a$(iii) $\sqrt3a^4+2\sqrt3a^3+3a^2-6a$ by $3a$To do:We have to divide the given expressions.Solution:We have to divide the given polynomials by monomials using the formula $x^a \div x^b=a^{a-b}$Polynomials: Polynomials are expressions in which each term is a constant multiplied by a variable raised to a whole number power.Monomial:A monomial is an expression that contains a single term composed of a product of constants and variables with non-negative integer exponents. Therefore, (i) The given expression is $-x^6+2x^4+4x^3+2x^2$ by $\sqrt2x^2$.$-x^6+2x^4+4x^3+2x^2 \div \sqrt2x^2=\frac{-x^6}{\sqrt2x^2}+\frac{2x^4}{\sqrt2x^2}+\frac{4x^3}{\sqrt2x^2}+\frac{2x^2}{\sqrt2x^2}$$-x^6+2x^4+4x^3+2x^2 \div \sqrt2x^2=\frac{-1}{\sqrt2}x^{6-2}+\frac{\sqrt2 \times \sqrt2}{\sqrt2}x^{4-2}+\frac{2\sqrt2 \times \sqrt2}{\sqrt2}x^{3-2}+\frac{\sqrt2 \times \sqrt2}{\sqrt2}x^{2-2}$$-x^6+2x^4+4x^3+2x^2 \div \sqrt2x^2=\frac{-1}{\sqrt2}x^{4}+\frac{\sqrt2}{1}x^{2}+\frac{2\sqrt2}{1}x^{1}+\frac{\sqrt2}{1}x^{0}$$-x^6+2x^4+4x^3+2x^2 \div \sqrt2x^2=\frac{-1}{\sqrt2}x^{4}+\sqrt2x^{2}+2\sqrt2x+\sqrt2$             [Since ... Read More