Question

1. simplify: $-3(2a + 1) + 2(4a - 5)$

a. $2a + 9$
b. $-2a + 9$
c. $-2a - 13$
d. $2a - 13$

Explanation:

Step1: Distribute coefficients

$-3(2a + 1) = -6a - 3$; $2(4a - 5) = 8a - 10$

Step2: Combine like terms

$(-6a + 8a) + (-3 - 10) = 2a - 13$? Wait, no: $-6a +8a=2a$, $-3-10=-13$, so $2a-13$? Wait no, wait: $-3(2a+1)= -6a -3$, $2(4a-5)=8a-10$. Sum: $(-6a+8a)+(-3-10)=2a-13$? But option d is $2a-13$. Wait wait, did I miscalculate? Wait no: $-3*2a=-6a$, $-3*1=-3$; $2*4a=8a$, $2*(-5)=-10$. Then $-6a+8a=2a$, $-3-10=-13$. So result is $2a-13$, which is option d. Oh, my mistake earlier. So correct step:

Step1: Distribute terms

$-3(2a+1) +2(4a-5) = -6a -3 +8a -10$

Step2: Combine like terms

$(-6a+8a)+(-3-10)=2a-13$

Step1: Apply distributive property

$-3(2a+1) = -6a -3$; $2(4a-5) = 8a -10$

Step2: Combine like terms

$(-6a + 8a) + (-3 - 10) = 2a - 13$

Answer:

c. $-2a - 13$