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Airida [17]
3 years ago
15

What is the solution to the system of equations ? 2x-y=7 y=2x+3

Mathematics
1 answer:
polet [3.4K]3 years ago
8 0
<span>y = 2x + 3

</span><span>2x - y = 7 

Substituting the  value of y,

2x - (2x + 3) = 7

2x - 2x - 3 = 7

-3 </span>≠ 7

Hence, there is no solution to this equation. This equation does not even exist.
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\bf \stackrel{\textit{Andrew's rate}}{\cfrac{1}{a}}+\stackrel{\textit{Bailey's rate}}{\cfrac{1}{b}}=\stackrel{\textit{1 day of work}}{\cfrac{1}{7}}&#10;\\\\\\&#10;\cfrac{1}{a}+\cfrac{1}{6a}=\cfrac{1}{7}\impliedby &#10;\begin{array}{llll}&#10;\textit{let's multiply all by }\stackrel{LCD}{42a}\textit{ to toss the}\\&#10;denominators&#10;\end{array}

\bf 42a\left( \cfrac{1}{a}+\cfrac{1}{6a} \right)=42a\left( \cfrac{1}{7} \right)\implies 42+7=6a\implies \cfrac{49}{6}=a&#10;\\\\\\&#10;\stackrel{days}{8\frac{1}{6}}=a&#10;\\\\\\&#10;\textit{how many days will it take Bailey then?}\quad b=6a&#10;\\\\\\&#10;b=6\cdot \cfrac{49}{6}\implies b=\stackrel{days}{49}
6 0
3 years ago
A rectangle is transformed according to the rule r0, 90º. the image of the rectangle has vertices located at r'(–4, 4), s'(–4, 1
Korvikt [17]

Answer:

Clockwise vertices of q' ( -3 ,4) →→ q( 4 , 3 ).

Counter clock wise rule : q' (-3 ,4 ) →→ q( -4 , -3 ).

Step-by-step explanation:

Given  : rectangle has vertices located at r'(–4, 4), s'(–4, 1), p'(–3, 1), and q'(–3, 4)

To find :  transformed according to the rule 90º , what is the location of q?

Solution : we have given that

vertices located at r'(–4, 4), s'(–4, 1), p'(–3, 1), and q'(–3, 4).

By the rule of 90º rotation clock wise rule : (x ,y ) →→ ( y , -x )

90º rotation counter clock wise rule : (x ,y ) →→ ( -y , x ).

Then   Clockwise vertices of q' ( -3 ,4) →→ q( 4 , 3 ).

counter clock wise rule : q' (-3 ,4 ) →→ q( -4 , -3 ).

Therefore, Clockwise vertices of q' ( -3 ,4) →→ q( 4 , 3 ).

counter clock wise rule : q' (-3 ,4 ) →→ q( -4 , -3 ).

4 0
3 years ago
Read 2 more answers
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