<u>Given </u><u>:</u><u>-</u>
<u>To </u><u>Find</u><u> </u><u>:</u><u>-</u>
- To convert the given equation into standard form.
<u>Solution</u><u> </u><u>:</u><u>-</u>
As we know that the standard form of the line is ,
ax + by + c = 0
So the given equation is ,
y = 2/5x -1/3
y = 6x - 5/15
15y = 6x - 5
6x -15y -5 = 0
<u>Hence</u><u> the</u><u> required</u><u> </u><u>equation</u><u> of</u><u> the</u><u> line</u><u> is</u><u> </u><u>6</u><u>x</u><u> </u><u>-15y-5=</u><u>0</u><u>.</u>
Hehe, I thought I'd get this one wrong, but x=10. Remember, that a straight angle must add up to 180 degrees. So I added all the angles; (3x+94)+(x+36)+(2x-4)=180. I took away the parentheses having 3x+94+x+36+2x-4=180. Remember from Algebra 1, to simplify terms, we add the polynomials. 94+36-4=126. 3x+x+2x= 6x. Our simplified equation is 120+6x=180. Now we can answer. 180-120=60. 60=6x. So therefore, for all x, x equals 10.
First you have multiply 6 by both 3x and 8
6*3x = 18x
6*8= 48
18x + 48 + 32 + 12x
now combine like terms
18x + 12x = 30x
48 + 32 = 80
so 30x + 80 is your answer
By algebra properties we find the following relationships between each pair of algebraic expressions:
- First equation: Case 4
- Second equation: Case 1
- Third equation: Case 2
- Fourth equation: Case 5
- Fifth equation: Case 3
<h3>How to determine pairs of equivalent equations</h3>
In this we must determine the equivalent algebraic expression related to given expressions, this can be done by applying algebra properties on equations from the second column until equivalent expression is found. Now we proceed to find for each case:
First equation
(7 - 2 · x) + (3 · x - 11)
(7 - 11) + (- 2 · x + 3 · x)
- 4 + (- 2 + 3) · x
- 4 + (1) · x
- 4 + (5 - 4) · x
- 4 - 4 · x + 5 · x
- 4 · (x + 1) + 5 · x → Case 4
Second equation
- 7 + 6 · x - 4 · x + 3
(6 · x - 4 · x) + (- 7 + 3)
(6 - 4) · x - 4
2 · x - 4
2 · (x - 2) → Case 1
Third equation
9 · x - 2 · (3 · x - 3)
9 · x - 6 · x + 6
3 · x + 6
(2 + 1) · x + (14 - 8)
[1 - (- 2)] · x + (14 - 8)
(x + 14) - (8 - 2 · x) → Case 2
Fourth equation
- 3 · x + 6 + 4 · x
x + 6
(5 - 4) · x + (7 - 1)
(7 + 5 · x) + (- 4 · x - 1) → Case 5
Fifth equation
- 2 · x + 9 + 5 · x + 6
3 · x + 15
3 · (x + 5) → Case 3
To learn more on algebraic equations: brainly.com/question/24875240
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Answer:

Step-by-step explanation:
Hello,
Touchdown is 7 points, field goals is 3 points
Let's note a the number of touchdowns and b the number of field goals, we can write
(1) a + b = 80
(2) 7a + 3b = 516
(2) - 3*(1) gives
7a + 3b - 3(a + b) = 516 - 3*80
<=> 7a + 3b -3a - 3b = 516 - 240 = 276
<=> 4a = 276
<=> 
And then from (1) b = 80 - 69 = 11
Hope this helps