Question 53·SAT Desmos Regression Practice·Algebra

For infinitely many solutions, the sides must agree for every $x$. Each side is a linear expression: expanding its products gives $x$ terms and plain numbers, but no $x^2$. Two linear expressions that agree at two different inputs agree everywhere. Choose $0$ and $1$. If the sides were different lines, they could meet at only one input.

Type $x_{1}=[0,1]$, then type the given equation with every $x$ changed to $x_{1}$ and $=$ changed to $\sim$. The regression fits $a$ so both inputs make the sides equal. Desmos shows $a=-5$ under PARAMETERS. Since the sides are linear and agree at two inputs, they agree for every $x$. The required value is $-5$. Choice A.

The coefficient of $x$ is the number multiplying $x$. The first product contributes $\frac{a-2}{3}\cdot9$; the subtracted product contributes $-(2a+1)$. Multiply and combine:

The two products on the right contribute $2(a+4)$ and $a-5$. Distribute and combine: $2(a+4)+(a-5)=2a+8+a-5=3a+3$. The $-41$ is a plain number, so it contributes no $x$ term.

For the $x$ terms to cancel for every $x$, their coefficients must match: $a-7=3a+3$. Subtract $a$: $-7=2a+3$. Subtract $3$: $-10=2a$. Divide by $2$: $a=-5$. Matching $x$ terms alone isn't enough; the plain numbers must match too.

The left-hand constant term comes from the parts without $x$. Distribute: $-2(a-2)+4(2a+1)=-2a+4+8a+4$. Combine: $=6a+8$. On the right, distribute: $a+4-2(a-5)-41=a+4-2a+10-41$. Combine: $=-a-27$. At $a=-5$, both constants equal $-22$: $6(-5)+8=-22$ and $-(-5)-27=-22$. Both the $x$-coefficients and constants match, so the equation has infinitely many solutions when $a=-5$. Choice A.