If the point (7, 3) is on the graph of an equation, which statement must be true?

Mathematics

1 answer:

slavikrds [6]2 years ago

6 0

Answer:

The values x = 7 and y = 3 make the equation true.

Step-by-step explanation:

hope this helps

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$\large\begin{array}{l}\\\\\\ \textsf{We need to find a proper value for b, so that if we add it to}\\\textsf{the left side of the equation, we get a perfect square.}\\\\ \textsf{So,}\\\\ \begin{array}{cccccccccc} \mathsf{a^2}&\!\!\!+\!\!\!&\mathsf{2}&\!\!\!\cdot\!\!\!&\mathsf{a}&\!\!\!\cdot\!\!\!&\mathsf{b}&\!\!\!+\!\!\!&\mathsf{b^2}&=\mathsf{(a+b)^2}\\\\ \mathsf{\downarrow}&&&\!\!\!\!\!&\downarrow&\!\!\!\!\!&\downarrow&&\downarrow\\\\ \mathsf{x^2}&\!\!\!+\!\!\!&\mathsf{2}&\!\!\!\cdot\!\!\!&\mathsf{x}&\!\!\!\cdot\!\!\!&\mathsf{1}&\!\!\!+\!\!\!&\mathsf{1^2}&=\mathsf{(x+1)^2} \end{array}\\\\\\ \textsf{Just by comparing the expressions above, we can conclude that}\\\textsf{b must be equal to 1, so we get a perfect square:}\\\\ \mathsf{(x+1)^2=x^2+2x+1} \end{array}$

$\large\begin{array}{l}\\\\\\ \textsf{Back to (i), adding 1 to both sides:}\\\\ \mathsf{x^2+2x+1=-1+1}\\\\ \boxed{\begin{array}{c} \mathsf{(x+1)^2=0} \end{array}} \end{array}$

$\large\begin{array}{l}\\\\\\ \textsf{If you want to solve this equation for x, you will find}\\\\ \mathsf{x+1=0}\\\\ \mathsf{x=-1}\quad\longleftarrow\quad\textsf{(solution)} \end{array}$

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$\large\begin{array}{l} \textsf{Any doubt? Please, comment below.}\\\\\\ \textsf{Best regards! :-)} \end{array}$

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