The most simplified version would be:
The answer is: " 91 " .
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→ " B = 91 " .
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Explanation:
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Given:
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" A + B = 180 " ;
"A = -2x + 115 " ; ↔ A = 115 − 2x ;
"B = - 6x + 169 " ; ↔ B = 169 − 6x ;
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METHOD 1)
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Solve for "x" ; and then plug the solved value for "x" into the expression given for "B" ; to solve for "B"
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(115 − 2x) + (169 − 6x) =
115 − 2x + 169 − 6x = ?
→ Combine the "like terms" ; as follows:
+ 115 + 169 = + 284 ;
− 2x − 6x = − 8x ;
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And rewrite as:
" − 8x + 284 " ;
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→ " - 8x + 284 = 180 " ;
Subtract: "284" from each side of the equation:
→ " - 8x + 284 − 284 = 180 − 284 " ;
to get:
→ " -8x = -104 ;
Divide EACH SIDE of the equation by "-8 " ;
to isolate "x" on one side of the equation; & to solve for "x" ;
→ -8x / -8 = -104/-8 ;
→ x = 13
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Now, to find the value of "B" :
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"B = - 6x + 169 " ; ↔ B = 169 − 6x ;
↔ B = 169 − 6x ;
= 169 − 6(13) ; ===========> Plug in our "solved value, "13", for "x" ;
= 169 − (78) ;
= 91 ;
B = " 91 " .
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The answer is: " 91 " .
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→ " B = 91 " .
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Now; let us check our answer:
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→ A + B = 180 ;
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Plug in our "solved answer" ; which is "91", for "B" ; as follows:
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→ A + 91 = ? 180? ;
↔ A = ? 180 − 91 ? ;
→ A = ? -89 ? Yes!
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→ " A = -2x + 115 " ; ↔ A = 115 − 2x ;
Plug in our solved value for "x"; which is: "13" ;
" A = 115 − 2x " ;
→ A = ? 115 − 2(13) ? ;
→ A = ? 115 − (26) ? ;
→ A = ? 29 ? Yes!
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METHOD 2)
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Given:
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" A + B = 180 " ;
"A = -2x + 115 " ; ↔ A = 115 − 2x ;
"B = - 6x + 169 " ; ↔ B = 169 − 6x ;
→ Solve for the value of "B" :
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A + B = 180 ;
→ B = 180 − A ;
→ B = 180 − (115 − 2x) ;
→ B = 180 − 1(115 − 2x) ; ==========> {Note the "implied value of "1" } ;
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Note the "distributive property" of multiplication:__________________________________________________ a(b + c) = ab + ac ; <u><em>AND</em></u>:
a(b − c) = ab − ac .________________________________________________________
Let us examine the following part of the problem:
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→ " − 1(115 − 2x) " ;
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→ " − 1(115 − 2x) " = (-1 * 115) − (-1 * 2x) ;
= -115 − (-2x) ;
= -115 + 2x ;
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So we can bring down the: " {"B = 180 " ...}" portion ;
→and rewrite:
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→ B = 180 − 115 + 2x ;
→ B = 65 + 2x ;
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Now; given: "B = - 6x + 169 " ; ↔ B = 169 − 6x ;
→ " B = 169 − 6x = 65 + 2x " ;
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→ " 169 − 6x = 65 + 2x "
Subtract "65" from each side of the equation; & Subtract "2x" from each side of the equation:
→ 169 − 6x − 65 − 2x = 65 + 2x − 65 − 2x ;
to get:
→ " - 8x + 104 = 0 " ;
Subtract "104" from each side of the equation:
→ " - 8x + 104 − 104 = 0 − 104 " ;
to get:
→ " - 8x = - 104 ;
Divide each side of the equation by "-8" ;
to isolate "x" on one side of the equation; & to solve for "x" ;
→ -8x / -8 = -104 / -8 ;
to get:
→ x = 13 ;
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Now, let us solve for: " B " ; → {for which this very question/problem asks!} ;
→ B = 65 + 2x ;
Plug in our solved value, " 13 ", for "x" ;
→ B = 65 + 2(13) ;
= 65 + (26) ;
→ B = " 91 " .
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Also, check our answer:
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Given: "B = - 6x + 169 " ; ↔ B = 169 − 6x = 91 ;
When "x = 13 " ; does: " B = 91 " ?
→ Plug in our "solved value" of " 13 " for "x" ;
→ to see if: "B = 91" ; (when "x = 13") ;
→ B = 169 − 6x ;
= 169 − 6(13) ;
= 169 − (78)______________________________________________________
→ B = " 91 " .
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I personally use y2-y1 over x2-x1 because it's simple. Plot two y values and two x values. The first ones you plotted are x1 and y1, the others are x2 y2. Now subtract y2-y1 and x2-x1. Afterwards, divide the y value by the x value, that will give you slope.
Infinite
x=-5y+3 is the same as x+5y=3
