Whakaoti i te (8x+1)(x-7)geq(2x-3)(4x+5)

$$( 8 x + 1 ) ( x - 7 ) \geq ( 2 x - 3 ) ( 4 x + 5 )$$

Whakaoti mō x

$$x\leq \frac{8}{53}$$

Ngā Upane Otinga

Whakamahia te āhuatanga tuaritanga hei whakarea te 8x+1 ki te x-7 ka whakakotahi i ngā kupu rite.

$$8x^{2}-55x-7\geq \left(2x-3\right)\left(4x+5\right)$$

Whakamahia te āhuatanga tuaritanga hei whakarea te 2x-3 ki te 4x+5 ka whakakotahi i ngā kupu rite.

$$8x^{2}-55x-7\geq 8x^{2}-2x-15$$

Tangohia te 8x^{2} mai i ngā taha e rua.

$$8x^{2}-55x-7-8x^{2}\geq -2x-15$$

Pahekotia te 8x^{2} me -8x^{2}, ka 0.

$$-55x-7\geq -2x-15$$

Me tāpiri te 2x ki ngā taha e rua.

$$-55x-7+2x\geq -15$$

Pahekotia te -55x me 2x, ka -53x.

$$-53x-7\geq -15$$

Me tāpiri te 7 ki ngā taha e rua.

$$-53x\geq -15+7$$

Tāpirihia te -15 ki te 7, ka -8.

$$-53x\geq -8$$

Whakawehea ngā taha e rua ki te -53. I te mea he tōraro a -53, ka huri te ahunga koreōrite.

$$x\leq \frac{-8}{-53}$$

Ka taea te hautanga \frac{-8}{-53} te whakamāmā ki te \frac{8}{53} mā te tango tahi i te tohu tōraro i te taurunga me te tauraro.

$$x\leq \frac{8}{53}$$