Answer:
[tex] \bold{( \underline{2x}^2+ \underline{2}x+1)-(7x^2+ \underline{2}x+ \underline{15})=-5x^2-14} \\ [/tex]
Step-by-step explanation:
To explain how to approach this question, I am going to place letters in the blank spaces:
[tex] \quad \sf{(ax^2+bx+c)-(dx^2+ex+f)=-5x^2-14} \\ [/tex]
First, compare the coefficients of the x² terms on both sides of the equation:
[tex] \implies\: \sf{a - d = -5a−d=−5} \\ [/tex]
Therefore, we need to choose any two numbers in place of a and d whose difference is -5:
[tex]\sf\implies \: a = 2 \: and \: d = 7 \: as \: 2 - 7 = -5 \\ [/tex]
Similarly, upon comparing the coefficients of the x terms, we can see that there is no x term on the right side of the equation. Therefore, we need to choose numbers in place of b and e for which the difference is zero. So b and e should be the same number:
[tex]\sf\implies \: b = 2 \: and \: e = 2 \: as \: 2 - 2 = 0\\ [/tex]
Finally, compare the constant term on both sides of the equation:
[tex]\implies\: \sf{c - f = -14c−f=−14} \\ [/tex]
Therefore, we need to choose any two numbers in place of c and f whose difference is -14:
[tex] \sf \implies \: {c = 1 \: and \: f = 15 \: as \: 1 - 15 = -14} \\ [/tex]
Plug in the numbers in place of the letters on the left side of the equation:
[tex] \implies \: \underline{ \bold{(2x^2+2x+1)-(7x^2+2x+15)=-5x^2-14}} \\ [/tex]
[tex] \\ \large \sf{\pmb{Checking} : } \\ [/tex]
[tex] \sf{Remove \: the \: brackets:} \\ [/tex]
[tex] \sf \implies \: {2x^2+2x+1-7x^2-2x-15=-5x^2-14} \\ [/tex]
[tex] \sf {Collect \: like \: terms:} \\ [/tex]
[tex] \sf \implies \: {2x^2-7x^2+2x-2x+1-15=-5x^2-14} \\ [/tex]
[tex] \sf{Combine \: like \: terms:} \\ [/tex]
[tex] \sf \implies \: {-5x^2-14=-5x^2-14} \\ \\ [/tex]
Hence proving that the left and right sides of the equation are the same.
