Question

Try to solve for x and y

Answer 1

x = 7

y = 8

Explanation:

Create two equations.

Equation 1

Both triangles are isosceles triangles since two of their sides are congruent. Therefore, the base angles of the top triangle are equal:

`\implies y^2-8=x^2+x`

Equation 2

According to the Vertical Angles Theorem, when two straight lines intersect, the opposite vertical angles are congruent. Therefore, the apexes of the triangles are equal. This means the sum of the base angles are equal:

`\implies (y^2-8)+(x^2+x)=8x+8x`

`\implies y^2-8+x^2+x=16x`

`\implies y^2-8+x^2+x-x^2=-x^2+16x`

`\implies y^2-8+x=-x^2+16x`

`\implies y^2-8+x-x=-x^2+16x-x`

`\implies y^2-8=-x^2+15x`

Substitute the first equation into the second equation and solve for x:

`\implies x^2+x=-x^2+15x`

`\implies x^2+x+x^2=-x^2+15x+x^2`

`\implies 2x^2+x=15x`

`\implies 2x^2+x-15x=15x-15x`

`\implies 2x^2-14x=0`

`\implies 2(x^2-7x)=0`

`\implies x^2-7x=0`

`\implies x(x-7)=0`

Therefore:

`\implies x=0, \quad x=7`

If x was zero, some of the angles would be zero, which is impossible.

Therefore, the only valid solution is x = 7.

Substitute the found value of x into the first equation and solve for y:

`\implies y^2-8=(7)^2+7`

`\implies y^2-8=49+7`

`\implies y^2-8=56`

`\implies y^2=64`

`\implies y=8`

Solution

• x = 7
• y = 8