Question

mr. diaz
classwork
math concepts
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hw 34

1. is 7 the solution of the equation $7x - 10 = 39$?

¿es $\frac{7}{2}$ la solución de la ecuación $7x - 10 = 39$?
èske $\frac{7}{2}$ solisyon ekwasyon an $7x - 10 = 39$ ?

1. find each sum or difference.

a. $(3x^2 - x - 6) - (-2x^2 + 2x - 7)$
b. $(5x^2 - 8x + 3) + (x^2 + 5x - 2)$

1. use inverse operation to solve each equation. check your answer.

usa la operación inversa para resolver cada ecuación. comprueba tu respuesta.
sèvi ak operasyon envès pou rezoud chak ekwasyon. tcheke repons ou an.
(1) $y + 8 = 4$
(2) $p - 10 = 12$

Explanation:

Question 1:

Step 1: Substitute \( x = \frac{7}{2} \) into the equation

Substitute \( x = \frac{7}{2} \) into \( 7x - 10 \):
\( 7\times\frac{7}{2}-10=\frac{49}{2}-10=\frac{49 - 20}{2}=\frac{29}{2}=14.5 \)

Step 2: Compare with the right - hand side

The right - hand side of the equation is \( 39 \). Since \( 14.5
eq39 \), \( \frac{7}{2} \) is not the solution of the equation \( 7x - 10 = 39 \).

Step 1: Remove the parentheses

Using the rule \( (a)-(b + c - d)=a - b - c + d \), for \( (3x^{2}-x - 6)-(-2x^{2}+2x - 7) \), we get:
\( 3x^{2}-x - 6 + 2x^{2}-2x + 7 \)

Step 2: Combine like terms

Combine the \( x^{2} \) terms: \( 3x^{2}+2x^{2}=5x^{2} \)
Combine the \( x \) terms: \( -x-2x=-3x \)
Combine the constant terms: \( -6 + 7 = 1 \)
So the result is \( 5x^{2}-3x + 1 \)

Step 1: Remove the parentheses

Using the rule \( (a)+(b + c - d)=a + b + c - d \), for \( (5x^{2}-8x + 3)+(x^{2}+5x - 2) \), we get:
\( 5x^{2}-8x + 3+x^{2}+5x - 2 \)

Step 2: Combine like terms

Combine the \( x^{2} \) terms: \( 5x^{2}+x^{2}=6x^{2} \)
Combine the \( x \) terms: \( -8x + 5x=-3x \)
Combine the constant terms: \( 3-2 = 1 \)
So the result is \( 6x^{2}-3x + 1 \)

Answer:

No, \( \frac{7}{2} \) is not the solution.

Question 2a: