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

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

x uchun yechish

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

Yechim qadamlari

8x+1 ga x-7 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

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

2x-3 ga 4x+5 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

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

Ikkala tarafdan 8x^{2} ni ayirish.

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

0 ni olish uchun 8x^{2} va -8x^{2} ni birlashtirish.

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

2x ni ikki tarafga qo’shing.

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

-53x ni olish uchun -55x va 2x ni birlashtirish.

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

7 ni ikki tarafga qo’shing.

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

-8 olish uchun -15 va 7'ni qo'shing.

$$-53x\geq -8$$

Ikki tarafini -53 ga bo‘ling. -53 manfiy boʻlgani uchun tengsizlikning yo‘nalishi o‘zgaradi.

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

Ikkala surat va maxrajdan manfiy belgini olib tashlash bilan \frac{-8}{-53} kasrini \frac{8}{53} ga soddalashtirish mumkin.

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