Examples:
An identity equation is an equation that is always true for any value substituted into the variable. For example, 2(x+1)=2x+22(x+1)=2x+2 is an identity equation. One way of checking is by simplifying the equation: \begin{aligned} 2(x+1)&=2x+2\\ 2x+2&=2x+2\\ 2&=2. \end{aligned} 2(x+1) 2x+2 2 ​ =2x+2 =2x+2 =2. ​ 2=22=2 is a true statement. Getting this kind of form is an indicator that the equation is in fact an identity equation. If we check by substituting different numbers, we see that the above assertion is indeed true. The following are identity equations: \begin{aligned} a(x+b)&=ax+ab\\ { (x+1) }^{ 2 }&={ x }^{ 2 }+2x+1\\ { (x+y) }^{ 2 }&={ x }^{ 2 }+2xy+{ y }^{ 2 }\\ { \sin }^{ 2 }\theta +{ \cos }^{ 2 }\theta &=1. \end{aligned} a(x+b) (x+1) 2 (x+y) 2 sin 2 θ+cos 2 θ ​ =ax+ab =x 2 +2x+1 =x 2 +2xy+y 2 =1. ​ The last equation is called a trigonometric identity. Solving identity equations: When given an identity equation in certain variables, start by collecting like terms (terms of the same variable and degree) together. Doing this will usually pair terms one on one, thus making it easier to solve. Let's see some examples: Given that (5x+3)-(2x+1)=ax+b(5x+3)−(2x+1)=ax+b is an algebraic identity in x,x, what are the values of aa and b?b? First, let us simplify the identity as follows: \begin{aligned} (5x+3)-(2x+1)&=ax+b\\ (5x-2x)+(3-1)&=ax+b\\ 3x+2&=ax+b. \end{aligned} (5x+3)−(2x+1) (5x−2x)+(3−1) 3x+2 ​ =ax+b =ax+b =ax+b. ​ Collecting like terms, we have \begin{aligned} 3x-ax+2-b&=0\\ x(3-a)+(2-b)&=0. \end{aligned} 3x−ax+2−b x(3−a)+(2−b) ​ =0 =0. ​ For the above identity to be true, both of the expressions on the left-hand side must be equal to zero. Thus we have 3-a=03−a=0 and 2-b=02−b=0, implying a=3, b=2. \ _\square a=3,b=2. □ ​ Given that ax^{3}+5y-cz+16=16x^{3}+by-3z+dax 3 +5y−cz+16=16x 3 +by−3z+d is an algebraic identity in x, y,x,y, and z,z, what are the values of a, b, ca,b,c and d?d? Since the identity is in terms of x, y,x,y, and zz, collect like terms with these variables: \begin{aligned} ax^{3}+5y-cz+16&=16x^{3}+by-3z+d\\ x^{3}(a-16)+y(5-b)-z(c-3)+(16-d)&=0. \end{aligned} ax 3 +5y−cz+16 x 3 (a−16)+y(5−b)−z(c−3)+(16−d) ​ =16x 3 +by−3z+d =0. ​ For the above equation to always be a true statement, that is 0=00=0, all the terms in the left side must be equal to 00. So we have a-16=0,\ 5-b=0,\ c-3=0,\ 16-d=0, a−16=0, 5−b=0, c−3=0, 16−d=0, implying a,b,c,da,b,c,d are equal to 16, 5, 3, 16,16,5,3,16, respectively. _\square □ ​ Given that (2x+ay)^{2}=bx^{2}+cxy+16y^{2}(2x+ay) 2 =bx 2 +cxy+16y 2 is an algebraic identity in x, y,x,y, and z,z, what are the value of a, ba,b and c?c? By the identity (x+y) ^{ 2 }={ x }^{ 2 }+2xy+y ^{ 2 }(x+y) 2 =x 2 +2xy+y 2 , the left side of the given identity is { (2x+ay) }^{ 2 }={ (2x) }^{ 2 }+2(2x)(ay)+{ (ay) }^{ 2 }. (2x+ay) 2 =(2x) 2 +2(2x)(ay)+(ay) 2 . Equating this with the right side gives, { 4x }^{ 2 }+4axy+{ { a }^{ 2 }y }^{ 2 }=b{ x }^{ 2 }+cxy+16{ y }^{ 2 }. 4x 2 +4axy+a 2 y 2 =bx 2 +cxy+16y 2 . Collecting like terms, we have { x }^{ 2 }(4-b)+xy(4a-c)+{ y }^{ 2 }({ a }^{ 2 }-16)=0. x 2 (4−b)+xy(4a−c)+y 2 (a 2 −16)=0. Making all the left terms zero to make the statement true, we have 4-b=0,\quad 4a-c=0,\quad { a }^{ 2 }-16=0, 4−b=0,4a−c=0,a 2 −16=0, which implies b=4, a=\pm 4, c=\pm 16. \ _\square b=4,a=±4,c=±16. □ ​ Given that a{ \sin }^{ 2 }\theta +a \cos^{ 2 }\theta =13asin 2 θ+acos 2 θ=13 is an algebraic identity in \theta,θ, what is the value of a?a? Using the above trigonometric identity { \sin }^{ 2 }\theta +{ \cos }^{ 2 }\theta =1,sin 2 θ+cos 2 θ=1, we have \begin{aligned} a{ \sin }^{ 2 }\theta +a\cos^{ 2 }\theta &=13\\ a({ \sin }^{ 2 }\theta +\cos^{ 2 }\theta) &=13\\ a\cdot 1&=13\\ a&=13. \ _\square \end{aligned} asin 2 θ+acos 2 θ a(sin 2 θ+cos 2 θ) a⋅1 a ​ =13 =13 =13 =13. □ ​ ​ Condition for an identity in x:x: If an equation in the form ax^2 + bx + cax 2 +bx+c has more than two values of xx satisfying the equation, then the condition is \color{#333333} a = b = c = 0. a=b=c=0. Find the value of rr in the equation (r^2 - 2r + 1)x^2 + (r^2 - 3r + 2)x - (r^2 + 2r - 3) = 0?(r 2 −2r+1)x 2 +(r 2 −3r+2)x−(r 2 +2r−3)=0? We will now use the above condition to solve the problem: Given: a = (r^2 - 2r + 1)x^2, b = (r^2 - 3r + 2), c = (r^2 + 2r -3)a=(r 2 −2r+1)x 2 ,b=(r 2 −3r+2),c=(r 2 +2r−3) Condition: a = b = c = 0a=b=c=0 r^2 - 2r + 1 = 0 \implies (r - 1)(r - 1) = 0 \implies r = 1, 1.r 2 −2r+1=0⟹(r−1)(r−1)=0⟹r=1,1. r^2 - 3r + 2 = 0 \implies (r - 2)(r - 1) = 0 \implies r = 2, 1.r 2 −3r+2=0⟹(r−2)(r−1)=0⟹r=2,1. r^2 + 2r - 3 = 0 \implies (r + 3)(r - 1) = 0 \implies r = -3, 1.r 2 +2r−3=0⟹(r+3)(r−1)=0⟹r=−3,1. Out of all the values, we must now find the common value for r,r, which is 1.
