Whakaoti i te 3/4[4/3(1/2x-1/4)-8]=frac{3}{2}x+1

$$\frac { 3 } { 4 } [ \frac { 4 } { 3 } ( \frac { 1 } { 2 } x - \frac { 1 } { 4 } ) - 8 ] = \frac { 3 } { 2 } x + 1$$

Whakaoti mō x

$$x = -\frac{29}{4} = -7\frac{1}{4} = -7.25$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakamahia te āhuatanga tohatoha hei whakarea te \frac{4}{3} ki te \frac{1}{2}x-\frac{1}{4}.

$$\frac{3}{4}\left(\frac{4}{3}\times \frac{1}{2}x+\frac{4}{3}\left(-\frac{1}{4}\right)-8\right)=\frac{3}{2}x+1$$

Me whakarea te \frac{4}{3} ki te \frac{1}{2} mā te whakarea taurunga ki te taurunga me te tauraro ki te tauraro.

$$\frac{3}{4}\left(\frac{4\times 1}{3\times 2}x+\frac{4}{3}\left(-\frac{1}{4}\right)-8\right)=\frac{3}{2}x+1$$

Mahia ngā whakarea i roto i te hautanga \frac{4\times 1}{3\times 2}.

$$\frac{3}{4}\left(\frac{4}{6}x+\frac{4}{3}\left(-\frac{1}{4}\right)-8\right)=\frac{3}{2}x+1$$

Whakahekea te hautanga \frac{4}{6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$\frac{3}{4}\left(\frac{2}{3}x+\frac{4}{3}\left(-\frac{1}{4}\right)-8\right)=\frac{3}{2}x+1$$

Me whakarea te \frac{4}{3} ki te -\frac{1}{4} mā te whakarea taurunga ki te taurunga me te tauraro ki te tauraro.

$$\frac{3}{4}\left(\frac{2}{3}x+\frac{4\left(-1\right)}{3\times 4}-8\right)=\frac{3}{2}x+1$$

Me whakakore tahi te 4 i te taurunga me te tauraro.

$$\frac{3}{4}\left(\frac{2}{3}x+\frac{-1}{3}-8\right)=\frac{3}{2}x+1$$

Ka taea te hautanga \frac{-1}{3} te tuhi anō ko -\frac{1}{3} mā te tango i te tohu tōraro.

$$\frac{3}{4}\left(\frac{2}{3}x-\frac{1}{3}-8\right)=\frac{3}{2}x+1$$

Me tahuri te 8 ki te hautau \frac{24}{3}.

$$\frac{3}{4}\left(\frac{2}{3}x-\frac{1}{3}-\frac{24}{3}\right)=\frac{3}{2}x+1$$

Tā te mea he rite te tauraro o -\frac{1}{3} me \frac{24}{3}, me tango rāua mā te tango i ō raua taurunga.

$$\frac{3}{4}\left(\frac{2}{3}x+\frac{-1-24}{3}\right)=\frac{3}{2}x+1$$

Tangohia te 24 i te -1, ka -25.

$$\frac{3}{4}\left(\frac{2}{3}x-\frac{25}{3}\right)=\frac{3}{2}x+1$$

Whakamahia te āhuatanga tohatoha hei whakarea te \frac{3}{4} ki te \frac{2}{3}x-\frac{25}{3}.

$$\frac{3}{4}\times \frac{2}{3}x+\frac{3}{4}\left(-\frac{25}{3}\right)=\frac{3}{2}x+1$$

Me whakarea te \frac{3}{4} ki te \frac{2}{3} mā te whakarea taurunga ki te taurunga me te tauraro ki te tauraro.

$$\frac{3\times 2}{4\times 3}x+\frac{3}{4}\left(-\frac{25}{3}\right)=\frac{3}{2}x+1$$

Me whakakore tahi te 3 i te taurunga me te tauraro.

$$\frac{2}{4}x+\frac{3}{4}\left(-\frac{25}{3}\right)=\frac{3}{2}x+1$$

Whakahekea te hautanga \frac{2}{4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$\frac{1}{2}x+\frac{3}{4}\left(-\frac{25}{3}\right)=\frac{3}{2}x+1$$

Me whakarea te \frac{3}{4} ki te -\frac{25}{3} mā te whakarea taurunga ki te taurunga me te tauraro ki te tauraro.

$$\frac{1}{2}x+\frac{3\left(-25\right)}{4\times 3}=\frac{3}{2}x+1$$

Me whakakore tahi te 3 i te taurunga me te tauraro.

$$\frac{1}{2}x+\frac{-25}{4}=\frac{3}{2}x+1$$

Ka taea te hautanga \frac{-25}{4} te tuhi anō ko -\frac{25}{4} mā te tango i te tohu tōraro.

$$\frac{1}{2}x-\frac{25}{4}=\frac{3}{2}x+1$$

Tangohia te \frac{3}{2}x mai i ngā taha e rua.

$$\frac{1}{2}x-\frac{25}{4}-\frac{3}{2}x=1$$

Pahekotia te \frac{1}{2}x me -\frac{3}{2}x, ka -x.

$$-x-\frac{25}{4}=1$$

Me tāpiri te \frac{25}{4} ki ngā taha e rua.

$$-x=1+\frac{25}{4}$$

Me tahuri te 1 ki te hautau \frac{4}{4}.

$$-x=\frac{4}{4}+\frac{25}{4}$$

Tā te mea he rite te tauraro o \frac{4}{4} me \frac{25}{4}, me tāpiri rāua mā te tāpiri i ō raua taurunga.

$$-x=\frac{4+25}{4}$$

Tāpirihia te 4 ki te 25, ka 29.

$$-x=\frac{29}{4}$$

Me whakarea ngā taha e rua ki te -1.

$$x=-\frac{29}{4}$$