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Found 225 Articles for Class 8
75 Views
Given:The given equations are:(i) $\frac{45-2x}{15}-\frac{4x+10}{5}=\frac{15-14x}{9}$(ii) $\frac{5(7x+5)}{3}-\frac{23}{3}=13-\frac{4x-2}{3}$To do:We have to solve the given equations and check the results.Solution:To check the results 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{45-2x}{15}-\frac{4x+10}{5}=\frac{15-14x}{9}$$\frac{45-2x}{15}-\frac{4x+10}{5}=\frac{15-14x}{9}$On rearranging, we get, $\frac{45-2x}{15}-\frac{4x+10}{5}-\frac{15-14x}{9}=0$LCM of denominators $15, 5$ and $9$ is $45$$\frac{(45-2x)\times3-(4x+10)\times9-(15-14x) \times5}{45}=0$$\frac{3(45)-3(2x)-9(4x)-9(10)-5(15)+5(14x)}{45}=0$$\frac{135-6x-36x-90-75+70x}{45}=0$$\frac{135-165-42x+70x}{45}=0$$\frac{-30+28x}{45}=0$On cross multiplication, we get, $28x-30=45(0)$$28x-30=0$$28x=30$$x=\frac{30}{28}$$x=\frac{15}{14}$Verification:LHS $=\frac{45-2x}{15}-\frac{4x+10}{5}$$=\frac{45-2(\frac{15}{14})}{15}-\frac{4(\frac{15}{14})+10}{5}$$=\frac{45-\frac{15}{7}}{15}-\frac{\frac{30}{7}+10}{5}$$=\frac{45\times7-15}{7\times15}-\frac{30+10\times7}{7\times5}$$=\frac{315-15}{105}-\frac{30+70}{35}$$=\frac{300}{105}-\frac{100}{35}$$=\frac{60}{21}-\frac{20}{7}$$=\frac{60-20\times3}{21}$$=\frac{60-60}{21}$$=0$RHS $=\frac{15-14x}{9}$$=\frac{15-14(\frac{15}{14})}{9}$$=\frac{15-15}{9}$$=0$LHS $=$ RHSHence verified.(ii) The given equation is $\frac{5(7x+5)}{3}-\frac{23}{3}=13-\frac{4x-2}{3}$$\frac{5(7x+5)}{3}-\frac{23}{3}=13-\frac{4x-2}{3}$On rearranging, we get, $\frac{5(7x+5)}{3}+\frac{4x-2}{3}=\frac{23}{3}+13$LCM of $3$ and $1$ is $3$$\frac{5(7x)+5(5)+4x-2}{3}=\frac{23+13\times3}{3}$$\frac{35x+25+4x-2}{3}=\frac{23+39}{3}$$\frac{39x+23}{3}=\frac{62}{3}$On cross multiplication, we get, $39x+23=62$$39x=62-23$$39x=39$$x=\frac{39}{39}$$x=1$Verification:LHS $=\frac{5(7x+5)}{3}-\frac{23}{3}$$=\frac{5(7(1)+5)}{3}-\frac{23}{3}$$=\frac{5(7+5)}{3}-\frac{23}{3}$$=\frac{5(12)}{3}-\frac{23}{3}$$=\frac{60}{3}-\frac{23}{3}$$=\frac{60-23}{3}$$=\frac{37}{3}$RHS $=13-\frac{4x-2}{3}$$=13-\frac{4(1)-2}{3}$$=13-\frac{4-2}{3}$$=13-\frac{2}{3}$$=\frac{13\times3-2}{3})$$=\frac{39-2}{3}$$=\frac{37}{3}$LHS $=$ RHSHence ... Read More
69 Views
Given:The given equations are:(i) $\frac{2}{3x}-\frac{3}{2x}=\frac{1}{12}$(ii) $\frac{4x}{9}+\frac{1}{3}+\frac{13x}{108}=\frac{8x+19}{18}$To do:We have to solve the given equations and check the results.Solution:To check the results 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{2}{3x}-\frac{3}{2x}=\frac{1}{12}$$\frac{2}{3x}-\frac{3}{2x}=\frac{1}{12}$LCM of denominators $3x$ and $2x$ is $6x$$\frac{2\times2-3\times3}{6x}=\frac{1}{12}$$\frac{4-9}{6x}=\frac{1}{12}$$\frac{-5}{6x}=\frac{1}{12}$On cross multiplication, we get, $-5\times12=1\times6x$$6x=-60$$x=\frac{-60}{6}$$x=-10$Verification:LHS $=\frac{2}{3x}-\frac{3}{2x}$$=\frac{2}{3(-10)}-\frac{3}{2(-10)}$$=\frac{2}{-30}-\frac{3}{-20}$$=\frac{-1}{15}-(\frac{-3}{20}$$=\frac{-1}{15}+\frac{3}{20}$$=\frac{-1\times4+3\times3}{60}$ (LCM of $15$ and $20$ is $60$)$=\frac{-4+9}{60}$$=\frac{5}{60}$$=\frac{1}{12}$RHS $=\frac{1}{12}$LHS $=$ RHSHence verified.(ii) The given equation is $\frac{4x}{9}+\frac{1}{3}+\frac{13x}{108}=\frac{8x+19}{18}$$\frac{4x}{9}+\frac{1}{3}+\frac{13x}{108}=\frac{8x+19}{18}$On rearranging, we get, $\frac{4x}{9}+\frac{13x}{108}-\frac{8x+19}{18}=-\frac{1}{3}$LCM of $9, 108$ and $18$ is $108$$\frac{4x \times 12+13x \times1- (8x+19)\times6}{108}=-\frac{1}{3}$$\frac{48x+13x-48x-114}{108}=-\frac{1}{3}$$\frac{13x-114}{108}=-\frac{1}{3}$On ... Read More
83 Views
Given:The given equations are:(i) $\frac{9x+7}{2}-(x-\frac{(x-2)}{7})=36$(ii) $0.18(5x-4)=0.5x+0.8$To do:We have to solve the given equations and check the results.Solution:To check the results 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{9x+7}{2}-(x-\frac{(x-2)}{7})=36$$\frac{9x+7}{2}-(x-\frac{(x-2)}{7})=36$$\frac{9x+7}{2}-(\frac{7x-(x-2)}{7})=36$$\frac{9x+7}{2}-(\frac{7x-x+2}{7})=36$$\frac{9x+7}{2}-(\frac{6x+2)}{7})=36$LCM of denominators $2$ and $7$ is $14$$\frac{(9x+7)\times7-(6x+2)\times2}{14}=36$$\frac{7(9x)+7(7)-2(6x)-2(2)}{14}=36$$\frac{63x+49-12x-4}{14}=36$$\frac{51x+45}{14}=36$On cross multiplication, we get, $51x+45=36\times14$$51x+45=504$$51x=504-45$$51x=459$$x=\frac{459}{51}$$x=9$Verification:LHS $=\frac{9x+7}{2}-(x-\frac{(x-2)}{7})$$=\frac{9(9)+7}{2}-(9-\frac{(9-2)}{7})$$=\frac{81+7}{2}-(9-\frac{7}{7})$$=\frac{88}{2}-(9-1)$$=44-8$$=36$RHS $=36$LHS $=$ RHSHence verified.(ii) The given equation is $0.18(5x-4)=0.5x+0.8$$0.18(5x-4)=0.5x+0.8$$0.18(5x)-0.18(4)=0.5x+0.8$$0.9x-0.72=0.5x+0.8$On rearranging, we get, $0.9x-0.5x=0.8+0.72$$0.4x=1.52$$x=\frac{1.52}{0.4}$$x=3.8$Verification:LHS $=0.18(5x-4)$$=0.18(5(3.8)-4)$$=0.18(19-4)$$=0.18(15)$$=2.7$RHS $=0.5x+0.8$$=0.5(3.8)+0.8$$=1.9+0.8$$=2.7$LHS $=$ RHSHence verified.Read More
73 Views
Given:The given equations are:(i) $\frac{3x+1}{16}+\frac{2x-3}{7}=\frac{x+3}{8}+\frac{3x-1}{14}$(ii) $\frac{1-2x}{7}-\frac{2-3x}{8}=\frac{3}{2}+\frac{x}{4}$To do:We have to solve the given equations and check the results.Solution:To check the results we have to find the values of the variables and substitute them in the equation. Find the value of LHS and the value of RHS and check whether both are equal.(i) The given equation is $\frac{3x+1}{16}+\frac{2x-3}{7}=\frac{x+3}{8}+\frac{3x-1}{14}$$\frac{3x+1}{16}+\frac{2x-3}{7}=\frac{x+3}{8}+\frac{3x-1}{14}$On rearranging, we get, $\frac{3x+1}{16}+\frac{2x-3}{7}-\frac{x+3}{8}-\frac{3x-1}{14}=0$LCM of denominators $16, 7, 8$ and $14$ is $112$$\frac{(3x+1)\times7+(2x-3)\times16-(x+3) \times14-(3x-1)\times8}{112}=0$$\frac{7(3x)+7(1)+16(2x)-16(3)-14(x)-14(3)-8(3x)+8(1)}{112}=0$$\frac{21x+7+32x-48-14x-42-24x+8}{112}=0$$\frac{53x-38x+15-90}{112}=0$$\frac{15x-75}{112}=0$On cross multiplication, we get, $15x-75=112(0)$$15x-75=0$$15x=75$$x=\frac{75}{15}$$x=5$Verification:LHS $=\frac{3x+1}{16}+\frac{2x-3}{7}$$=\frac{3(5)+1}{16}+\frac{2(5)-3}{7}$$=\frac{15+1}{16}+\frac{10-3}{7}$$=\frac{16}{16}+\frac{7}{7}$$=1+1$$=2$RHS $=\frac{x+3}{8}+\frac{3x-1}{14}$$=\frac{5+3}{8}+\frac{3(5)-1}{14}$$=\frac{8}{8}+\frac{15-1}{14}$$=1+\frac{14}{14}$$=1+1$$=2$LHS $=$ RHSHence verified.(ii) The given equation is $\frac{1-2x}{7}-\frac{2-3x}{8}=\frac{3}{2}+\frac{x}{4}$$\frac{1-2x}{7}-\frac{2-3x}{8}=\frac{3}{2}+\frac{x}{4}$On rearranging, we get, $\frac{1-2x}{7}-\frac{2-3x}{8}-\frac{x}{4}=\frac{3}{2}$LCM of $7, 8$ and $4$ is $56$$\frac{8\times (1-2x)-(2-3x)\times7-(x)\times14}{56}=\frac{3}{2}$$\frac{8-16x-14+21x-14x}{56}=\frac{3}{2}$$\frac{-9x-6}{56}=\frac{3}{2}$On cross multiplication, we get, $(-9x-6)\times2=3\times56$$-18x-12=168$$-18x=168+12$$-18x=180$$x=\frac{180}{-18}$$x=-10$Verification:LHS $=\frac{1-2x}{7}-\frac{2-3x}{8}$$=\frac{1-2(-10)}{7}-\frac{2-3(-10)}{8}$$=\frac{1+20}{7}-\frac{2+30}{8}$$=\frac{21}{7}-\frac{32}{8}$$=3-4$$=-1$RHS ... Read More
96 Views
Given:The given equations are:(i) $\frac{3x}{4}-\frac{x-1}{2}=\frac{x-2}{3}$(ii) $\frac{5x}{3}-\frac{(x-1)}{4}=\frac{(x-3)}{5}$To do:We have to solve the given equations and check the results.Solution:To check the results we have to find the values of the variables and substitute them in the equation. Find the value of LHS and the value of RHS and check whether both are equal.(i) The given equation is $\frac{3x}{4}-\frac{x-1}{2}=\frac{x-2}{3}$$\frac{3x}{4}-\frac{x-1}{2}=\frac{x-2}{3}$On rearranging, we get, $\frac{3x}{4}-\frac{x-1}{2}-\frac{x-2}{3}=0$LCM of denominators $4, 2$ and $3$ is $12$$\frac{(3x)\times3-(x-1)\times6-(x-2) \times4}{4}=0$$\frac{9x-6(x)+6(1)-4(x)+4(2)}{12}=0$$\frac{9x-6x+6-4x+8}{12}=0$$\frac{-x+14}{12}=0$On cross multiplication, we get, $-x+14=0(12)$$-x+14=0$$x=14$Verification:LHS $=\frac{3x}{4}-\frac{x-1}{2}$$=\frac{3(14)}{4}-\frac{14-1}{2}$$=\frac{42}{4}-\frac{13}{2}$$=\frac{21}{2}-\frac{13}{2}$$=\frac{21-13}{2}$$=\frac{8}{2}$$=4$RHS $=\frac{x-2}{3}$$=\frac{14-2}{3}$$=\frac{12}{3}$$=4$LHS $=$ RHSHence verified.(ii) The given equation is $\frac{5x}{3}-\frac{(x-1)}{4}=\frac{(x-3)}{5}$.$\frac{5x}{3}-\frac{(x-1)}{4}=\frac{(x-3)}{5}$On rearranging, we get, $\frac{5x}{3}-\frac{(x-1)}{4}-\frac{(x-3)}{5}=0$LCM of $3, 4$ and $5$ is $60$$\frac{5x \times 20-(x-1)\times15-(x-3)\times12}{60}=0$$\frac{100x-15x+15-12x+36}{60}=0$$\frac{73x+51}{60}=0$On cross multiplication, we get, $73x+51=60(0)$$73x+51=0$$73x=-51$$x=\frac{-51}{73}$Verification:LHS $=\frac{5x}{3}-\frac{(x-1)}{4}$$=\frac{5(\frac{-51}{73})}{3}-\frac{(\frac{-51}{73}-1)}{4}$$=\frac{\frac{5\times(-51)}{73}}{3}-\frac{\frac{-51-1\times73}{73}}{4}$$=\frac{-255}{219}-\frac{-51-73}{73\times4}$$=\frac{-255}{219}-\frac{-124}{292}$$=\frac{-255\times4+124\times3}{876}$ ... Read More
82 Views
Given:The given equations are:(i) $\frac{(3a-2)}{3}+\frac{(2a+3)}{2}=a+\frac{7}{6}$(ii) $x-\frac{(x-1)}{2}=1-\frac{(x-2)}{3}$To do:We have to solve the given equations and check the results.Solution:To check the results 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{(3a-2)}{3}+\frac{(2a+3)}{2}=a+\frac{7}{6}$$\frac{(3a-2)}{3}+\frac{(2a+3)}{2}=a+\frac{7}{6}$On rearranging, we get, $\frac{(3a-2)}{3}+\frac{(2a+3)}{2}-a=\frac{7}{6}$LCM of denominators $3$ and $2$ is $6$$\frac{(3a-2)\times2+(2a+3)\times3-a \times6}{6}=\frac{7}{6}$$\frac{2(3a)-2(2)+(2a(3)+3(3)-6a}{6}=\frac{7}{6}$$\frac{6a-4+6a+9-6a}{6}=\frac{7}{6}$$\frac{6a-4+9}{6}=\frac{7}{6}$$\frac{6a+5}{6}=\frac{7}{6}$On cross multiplication, we get, $6a+5=\frac{7\times6}{6}$$6a+5=7$$6a+5=7$$6a=7-5$$6a=2$$a=\frac{2}{6}$$a=\frac{1}{3}$Verification:LHS $=\frac{(3a-2)}{3}+\frac{(2a+3)}{2}$$=\frac{(3(\frac{1}{3})-2)}{3}+\frac{(2(\frac{1}{3})+3)}{2}$$=\frac{1-2}{3}+\frac{\frac{2}{3}+3}{2}$$=\frac{-1}{3}+\frac{\frac{2+3\times3}{3}}{2}$$=\frac{-1}{3}+\frac{\frac{2+9}{3}}{2}$$=\frac{-1}{3}+\frac{11}{3\times2}$$=\frac{-1}{3}+\frac{11}{6}$$=\frac{-1\times2+11}{6}$ (LCM of $3$ and $6$ is $6$)$=\frac{-2+11}{6}$$=\frac{9}{6}$$=\frac{3}{2}$RHS $=a+\frac{7}{6}$$=\frac{1}{3}+\frac{7}{6}$$=\frac{1\times2+7}{6}$ (LCM of $3$ and $6$ is $6$)$=\frac{2+7}{6}$$=\frac{9}{6}$$=\frac{3}{2}$LHS $=$ ... Read More
57 Views
Given:The given equations are:(i) $\frac{7x}{2}-\frac{5x}{2}=\frac{20x}{3}+10$(ii) $\frac{6x+1}{2}+1=\frac{7x-3}{3}$To do:We have to solve the given equations and check the results.Solution:To check the results we have to find the values of the variables and substitute them in the equation. Find the value of LHS and the value of RHS and check whether both are equal.(i) The given equation is $\frac{7x}{2}-\frac{5x}{2}=\frac{20x}{3}+10$.$\frac{7x}{2}-\frac{5x}{2}=\frac{20x}{3}+10$On rearranging, we get, $\frac{7x}{2}-\frac{5x}{2}-\frac{20x}{3}=10$LCM of $2$ and $3$ is $6$$\frac{7x \times3-5x \times 3-20x \times2}{6}=10$$\frac{21x-15x-40x}{6}=10$$\frac{21x-55x}{6}=10$$\frac{-34x}{6}=10$$\frac{-17x}{3}=10$On cross multiplication, we get, $-17x=3(10)$$-17x=30$$x=\frac{30}{-17}$$x=\frac{-30}{17}$Verification:LHS $=\frac{7x}{2}-\frac{5x}{2}$$=\frac{7(\frac{-30}{17})}{2}-\frac{5(\frac{-30}{17})}{2}$$=\frac{-210}{34}-\frac{-150}{34}$$=\frac{-210+150}{34}$$=\frac{-60}{34}$$=\frac{-30}{17}$RHS $=\frac{20x}{3}+10$$=\frac{20(\frac{-30}{17})}{3}+10$$=\frac{20\times(-30)}{17\times3}+10$$=\frac{-600}{51}+10$$=\frac{-600+51\times10}{51}$ (LCM of $51$ and $1$ is $51$)$=\frac{-600+510}{51}$$=\frac{-90}{51}$$=\frac{-30}{17}$LHS $=$ RHSHence verified.(ii) The given equation is $\frac{6x+1}{2}+1=\frac{7x-3}{3}$$\frac{6x+1}{2}+1=\frac{7x-3}{3}$On ... Read More
68 Views
Given:The given equations are:(i) $\frac{1}{2}x+7x-6=7x+\frac{1}{4}$(ii) $\frac{3}{4}x+4x=\frac{7}{8}+6x-6$To do:We have to solve the given equations and check the results.Solution:To check the results we have to find the values of the variables and substitute them in the equation. Find the value of LHS and the value of RHS and check whether both are equal.(i) The given equation is $\frac{1}{2}x+7x-6=7x+\frac{1}{4}$.$\frac{1}{2}x+7x-6=7x+\frac{1}{4}$On rearranging, we get, $\frac{1}{2}x+7x-7x=\frac{1}{4}+6$$\frac{1}{2}x=\frac{1+6\times4}{4}$ (LCM of $4$ and $1$ is $4$)$\frac{1}{2}x=\frac{1+24}{4}$$\frac{1}{2}x=\frac{25}{4}$On cross multiplication, we get, $x=\frac{25\times2}{4}$$x=\frac{25}{2}$Verification:LHS $=\frac{1}{2}x+7x-6$$=\frac{1}{2}(\frac{25}{2})+7(\frac{25}{2})-6$$=\frac{25}{2\times2}+\frac{25\times7}{2}-6$$=\frac{25}{4}+\frac{175}{2}-6$$=\frac{25+175\times2-6\times4}{4}$ (LCM of $2$ and $4$ is $4$)$=\frac{25+350-24}{4}$$=\frac{351}{4}$RHS $=7x+\frac{1}{4}$$=7(\frac{25}{2})+\frac{1}{4}$$=\frac{25\times7}{2}+\frac{1}{4}$$=\frac{175}{2}+\frac{1}{4}$ $=\frac{175\times2+1}{4}$ ... Read More
65 Views
Given:The given equations are:(i) $\frac{7y+2}{5}=\frac{6y-5}{11}$(ii) $x-2x+2-\frac{16}{3}x+5=3-\frac{7}{2}x$To do:We have to solve the given equations and check the results.Solution:To check the results 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 $\frac{7y+2}{5}=\frac{6y-5}{11}$.$\frac{7y+2}{5}=\frac{6y-5}{11}$On cross multiplication, we get, $(7y+2)\times11=5(6y-5)$$11(7y)+11(2)=5(6y)-5(5)$$77y+22=30y-25$$77y-30y=-25-22$$47y=-47$$y=\frac{-47}{47}$$y=-1$Verification:LHS $=\frac{7y+2}{5}$$=\frac{7(-1)+2}{5}$$=\frac{-7+2}{5}$$=\frac{-5}{5}$$=-1$RHS $=\frac{6y-5}{11}$$=\frac{6(-1)-5}{11}$$=\frac{-6-5}{11}$$=\frac{-11}{11}$$=-1$LHS $=$ RHSHence verified.(ii) The given equation is $x-2x+2-\frac{16}{3}x+5=3-\frac{7}{2}x$$x-2x+2-\frac{16}{3}x+5=3-\frac{7}{2x}$On rearranging, we get, $x-2x-\frac{16}{3}x+\frac{7}{2}x=3-2-5$$-x-\frac{16}{3}x+\frac{7}{2}x=3-7$$x(-1-\frac{16}{3}+\frac{7}{2})=-4$LCM of denominators $3$ and $2$ is $6$$x(\frac{-1\times6-16\times2+7\times3}{6})=-4$$x(\frac{-6-32+21}{6})=-4$$x(\frac{-38+21}{6})=-4$$x(\frac{-17}{6})=-4$On cross multiplication, we get, $-17x=(-4)\times6$$-17x=-24$$x=\frac{-24}{-17}$$x=\frac{24}{17}$Verification:LHS $=x-2x+2-\frac{16}{3}x+5$$=\frac{24}{17}-2(\frac{24}{17})+2-\frac{16}{3}(\frac{24}{17})+5$$=\frac{24}{17}-\frac{48}{17}+2-\frac{16\times24}{3\times17}+5$$=\frac{24-48}{17}+7-\frac{16\times8}{17}$$=\frac{-24}{17}-\frac{128}{17}+7$$=\frac{-24-128+7\times17}{17}$$=\frac{-152+119}{17}$$=\frac{-33}{17}$RHS $=3-\frac{7}{2}x$$=3-\frac{7}{2}(\frac{24}{17})$$=3-\frac{7\times24}{2\times17}$$=3-\frac{7\times12}{17}$$=\frac{3\times17-84}{17}$$=\frac{51-84}{17}$$=\frac{-33}{17}$LHS $=$ RHSHence verified.Read More
61 Views
Given:The given equations are:(i) $\frac{2x+5}{3}=3x-10$(ii) $\frac{a-8}{3}=\frac{a-3}{2}$To do:We have to solve the given equations and check the results.Solution:To check the results we have to find the values of the variables and substitute them in the equation. Find the value of LHS and the value of RHS and check whether both are equal.(i) The given equation is $\frac{2x+5}{3}=3x-10$.$\frac{2x+5}{3}=3x-10$On cross multiplication, we get, $2x+5=3(3x-10)$$2x+5=3(3x)-3(10)$$2x+5=9x-30$$9x-2x=5+30$$7x=35$$x=\frac{35}{7}$$x=5$Verification:LHS $=\frac{2x+5}{3}$$=\frac{2\times5+5}{3}$$=\frac{10+5}{3}$$=\frac{15}{3}$$=5$RHS $=3x-10$$=3(5)-10$$=15-10$$=5$LHS $=$ RHSHence verified.(ii) The given equation is $\frac{a-8}{3}=\frac{a-3}{2}$$\frac{a-8}{3}=\frac{a-3}{2}$On cross multiplication, we get, $(a-8)\times2=(a-3)\times3$$a(2)-8(2)=a(3)-3(3)$$2a-16=3a-9$$3a-2a=9-16$$a=-7$Verification:LHS $=\frac{a-8}{3}$$=\frac{-7-8}{3}$$=\frac{-15}{3}$$=-5$RHS $=\frac{a-3}{2}$$=\frac{-7-3}{2}$$=\frac{-10}{2}$$=-5$LHS $=$ RHSHence verified.Read More