All categories

3 years ago

How to solve - 10+5x=7x-4

1 answer:

8 0

10+5x=7x-4
1. simplify 7x - 5x
10=2x-4

2. add four to both sides
10+4=2x
+ 4 4
14=2x

3. divide 14 to both sides
14=2x

x=7

You might be interested in

What would be the amount of his last monthly bill in order. The bill amount to be 110 for all 12 months

meriva

Add up all the numbers the divide by the amount of numbers in the set.
1199/11 = 109
B. 109 answer

8 0

3 years ago

Answers please! I’m a bit confused

scoray [572]

Answer:

all work is shown and pictured

5 0

3 years ago

Read 2 more answers

Y= 2x + 6 for x = 0,1,2,3,4

algol [13]

x=0 y=6

x=1 y=8

x=2 y=10

x=3 y=12

x=4 y=14

8 0

3 years ago

Y – 11 = 3(x – 2) in standard form

emmainna [20.7K]

Answer:

3x + y = -5

General Formulas and Concepts:

Pre-Algebra

Distributive Property
Equality Properties

Algebra I

Standard Form: Ax + By = C

Point-Slope Form: y - y₁ = m(x - x₁)

x₁ - x coordinate
y₁ - y coordinate
m - slope

Step-by-step explanation:

Step 1: Define

[PS] y - 11 = 3(x - 2)

Step 2: Rewrite

Find Standard Form

Distribute 3: y - 11 = 3x - 6
Subtract 3x on both sides: -3x - y - 11 = -6
Add 11 to both sides: -3x - y = 5
Factor -1: -1(3x + y) = 5
Divide -1 on both sides: 3x + y = -5

4 0

3 years ago

Which equation has the solutions x=1+or-square root of 5?

stiv31 [10]

We will proceed to solve each case to determine the solution of the problem.

case a) $x^{2}+2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=-4+1$

$x^{2}+2x+1=-3$

Rewrite as perfect squares

$(x+1)^{2}=-3$

$(x+1)=(+/-)\sqrt{-3}\\(x+1)=(+/-)\sqrt{3}i\\x=-1(+/-)\sqrt{3}i$

therefore

case a) is not the solution of the problem

case b) $x^{2}-2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=-4+1$

$x^{2}-2x+1=-3$

Rewrite as perfect squares

$(x-1)^{2}=-3$

$(x-1)=(+/-)\sqrt{-3}\\(x-1)=(+/-)\sqrt{3}i\\x=1(+/-)\sqrt{3}i$

therefore

case b) is not the solution of the problem

case c) $x^{2}+2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=4+1$

$x^{2}+2x+1=5$

Rewrite as perfect squares

$(x+1)^{2}=5$

$(x+1)=(+/-)\sqrt{5}\\x=-1(+/-)\sqrt{5}$

therefore

case c) is not the solution of the problem

case d) $x^{2}-2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=4+1$

$x^{2}-2x+1=5$

Rewrite as perfect squares

$(x-1)^{2}=5$

$(x-1)=(+/-)\sqrt{5}\\x=1(+/-)\sqrt{5}$

therefore

case d) is the solution of the problem

therefore

the answer is

$x^{2}-2x-4=0$

5 0

3 years ago

Read 2 more answers