Write a two-column proof. Given: 7y = 8x - 14; y = 6. Prove that : x = 7.

Mathematics

2 answers:

svet-max [94.6K]2 years ago

4 0

Answer:

Given: $7y =8x-14$ , y =6

To prove: x =7

On substituting the value of y =6 in the equation 7y = 8x -14 to solve for x.

$7(6)= 8x - 14$

42 = 8x -14 ......[1]

Addition Property of equality states that you add the same number to both sides of an equation.

Add 14 to both sides in equation [1];

42+14 =8x -14+14

Simplify:

56 = 8x

8x =56 ......[2]

Division Property of equality states that you divide the same number to both sides of an equation.

Divide by 8 to both sides of an equation [2];

$\frac{8x}{8} =\frac{56}{8}$

Simplify:

x =7 Hence proved!

A two Column proof:

Statement Reason

1. 7y = 8x -14 Given

y= 6

2. 7(6) = 8x -14 Substitution property

3. 42+14 = 8x Addition property of equality

4. 56 =8x Simplify

5. x =7 Division Property of equality

trasher [3.6K]2 years ago

4 0

Given - 7y=8x-14 : y=6

Substitution Property - 7(6)=8x-14

Distributive Property - 8x-14=42

Addition Property - 8x=56

Division Property - x=7

You might be interested in

Find two unit vectors orthogonal to both given vectors. i j k, 4i k

Maurinko [17]

The cross product of two vectors gives a third vector $\mathbf v$ that is orthogonal to the first two.

$\mathbf v=(\vec i+\vec j+\vec k)\times(4\,\vec i+\vec k)=\begin{vmatrix}\vec i&\vec j&\vec k\\1&1&1\\4&0&1\end{vmatrix}=\vec i+3\,\vec j-4\,\vec k$

Normalize this vector by dividing it by its norm:

$\dfrac{\mathbf v}{\|\mathbf v\|}=\dfrac{\vec i+3\,\vec j-4\,\vec k}{\sqrt{1^2+3^2+(-4)^2}}=\dfrac1{\sqrt{26}}(\vec i+3\,\vec j-4\vec k)$

To get another vector orthogonal to the first two, you can just change the sign and use $-\mathbf v$ .